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NCERT solutions for class 12 physics chapter 8 Electromagnetic Waves

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NCERT solutions for class 12 physics chapter 8 Electromagnetic Waves

Edited By Vishal kumar | Updated on Sep 11, 2023 09:08 AM IST | #CBSE Class 12th

NCERT Solutions for Class 12 Physics Chapter 8 –Download Free PDF

NCERT Solutions for Class 12 Physics Chapter 8 Electromagnetic Waves- If you're a Class 12 student concerned about NCERT solutions electromagnetic waves, you're in the right place on Careers360. On this NCERT solutions page, you'll find answers to all the exercises and additional exercise questions from one to fifteen. These class 12 chapter 8 physics NCERT solutions, authored by subject experts, are detailed and presented in easy-to-understand language

The Electromagnetic Waves NCERT solutions cover the question related to the concept of displacement current. Problems related to the Electromagnetic Spectrum are also discussed in NCERT Solutions for Class 12 Physics chapter 8 Electromagnetic Waves. The class 12 physics ch 8 NCERT solutions help in testing the knowledge about the concepts studied in a chapter. Students must refer to these solutions of NCERT to prepare for the examination effectively. The NCERT Solutions for Class 12 Physics Chapter 8 PDF download also helps in preparing for various entrance exams like JEE, NEET, etc.

The NCERT textbook is usually explicitly referenced in the questions that are asked in the CBSE EMW class 12 exams. One of the subjects that comes up frequently on the board test is electromagnetism. Therefore, it is advised that students use the electromagnetic waves NCERT solutions to get a firm understanding of both the chapter and the subject.

Free download electromagnetic waves class 12 ncert solutions PDF for CBSE exam.

NCERT Solutions for Class 12 Physics Chapter 8 Electromagnetic Waves

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NCERT Solutions for Class 12 Physics Chapter 8 Electromagnetic Waves: Exercise Question and Answer

1.(a) Figure 8.6 shows a capacitor made of two circular plates each of
radius 12 cm, and separated by 5.0 cm. The capacitor is being
charged by an external source (not shown in the figure). The
charging current is constant and equal to 0.15A.

Calculate the capacitance and the rate of change of potential
difference between the plates.

1594195598206

Answer:

Radius of the discs(r) = 12cm

Area of the discs(A) = \pi r^{2}=\pi (0.12)^{2}=.045m^{2}

Permittivity, \epsilon _{0} = 8.85\times 10^{-12}C^{2}N^{-1}m^{2}

Distance between the two discs = 5cm=0.05m

Capacitance= \frac{\varepsilon _{0}A}{d}

= \frac{8.85\times 10^{-12}C^{2}N^{-1}m^{2}\times .045m^{2}}{0.05m}

= 8.003\times 10^{-12}F

=8.003 pF

Rate\ of \ change\ of \ potential =\frac{dv}{dt}

But

V =\frac{Q}{C}

Therefore,

Rate\ of \ change\ of \ potential=\frac{d(\frac{Q}{C})}{dt}=\frac{dq}{Cdt}=\frac{i}{C}

=\frac{0.15A}{8.003pF}

= 1.87\times 10^{10}Vs^{-1}

1(b). Figure 8.6 shows a capacitor made of two circular plates each of radius 12 cm, and separated by 5.0 cm. The capacitor is being charged by an external source (not shown in the figure). The
charging current is constant and equal to 0.15A.

Obtain the displacement current across the plates.

1594195608735

Answer:

The value of the displacement current will be the same as that of the conduction current which is given to be 0.15A. Therefore displacement current is also 0.15A.

1.(c) Figure 8.6 shows a capacitor made of two circular plates each of radius 12 cm, and separated by 5.0 cm. The capacitor is being charged by an external source (not shown in the figure). The
charging current is constant and equal to 0.15A.

Is Kirchhoff’s first rule (junction rule) valid at each plate of the
capacitor? Explain.

1594195616698

Answer:

Yes, Kirchoff's First rule (junction rule) is valid at each plate of the capacitor. This might not seem like the case at first but once we take into consideration both the conduction and displacement current the Kirchoff's first rule will hold good.

2.(a) A parallel plate capacitor (Fig. 8.7) made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 V ac supply with a (angular) frequency of 300 rad s ^{-1} .

What is the rms value of the conduction current?

1594195911624

Answer:

Capacitance(C) of the parallel plate capacitor = 100 pF

Voltage(V) = 230 V

Angular Frequency (ω) = 300\ rad\ s^{-1}

Rms\ Current(I) =\frac{Voltage}{Capacitive\ Reactance}

Capacitive \ Reactance(X_{c})= \frac{1}{C\omega }=\frac{1}{100\ pF\times 300\ rad\ s^{-1}}

X_{c}= 3.33\times 10^{7} \Omega

I=\frac{V}{X_{c}}=\frac{230V}{3.33\times 10^{7}\Omega }= 6.9\times 10^{-6}A

RMS value of conduction current is 6.9\mu A

2.(b) A parallel plate capacitor (Fig. 8.7) made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 V ac supply with a (angular) frequency of 300 rad s ^{-1}

Is the conduction current equal to the displacement current?

1594195919158


Answer:

Yes, conduction current is equal to the displacement current. This will be the case because otherwise, we will get different values of the magnetic field at the same point by taking two different surfaces and applying Ampere – Maxwell Law.

2.(c) A parallel plate capacitor (Fig. 8.7) made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 V ac supply with a (angular) frequency of 300 rad\: \: s ^{-1}

Determine the amplitude of B at a point 3.0 cm from the axis between the plates.

1594195925850

Answer:

We know the Ampere - Maxwells Law,

\oint B\cdot \vec{dl} = \mu _{0}(i_{c}+\varepsilon _{0}\frac{d\phi _{E}}{dt})

Between the plates conduction current i_{c}=0 .

For a loop of radius r smaller than the radius of the discs,

\mu _{0}\varepsilon _{0}\frac{\mathrm{d} \phi _{E}}{\mathrm{d} t}=\mu _{0}i_{d}\frac{\pi r^{2}}{\pi R^{2}}=\mu _{0}i_{d}\frac{ r^{2}}{ R^{2}}

B(2\pi r)=\mu _{0}i_{d}\frac{r^{2}}{R^{2}}

B=\mu _{0}i_{d}\frac{r}{2\pi R^{2}}

Since we have to find the amplitude of the magnetic field we won't use the RMS value but the maximum value of current.

\\i_{max}=\sqrt{2}\times i_{rms}\\ i_{max}=\sqrt{2}\times6.9\mu A\\ i_{max}=9.76\mu A

\\B_{amp}=\mu _{0}i_{max}\frac{r}{2\pi R^{2}}\\ B_{amp}=\frac{4\pi\times 10^{-7}\times 9.33\times 10^{-6}\times (.03) }{2\pi\times (0.06)^{2} }\\ B_{amp}=1.63\times 10^{-11}T

The amplitude of B at a point 3.0 cm from the axis between the plates is 1.63\times 10^{-11}T .

3. What physical quantity is the same for X-rays of wavelength 10 ^{-10} m, red light of wavelength 6800 \dot{A} and radiowaves of wavelength 500m?

Answer:

The speed with which these electromagnetic waves travel in a vacuum will be the same and will be equal to 3\times 10^{8}ms^{-1} (speed of light in vacuum).

4. A plane electromagnetic wave travels in vacuum along z-direction. What can you say about the directions of its electric and magnetic field vectors? If the frequency of the wave is 30 MHz, what is its wavelength?

Answer:

Since the electromagnetic wave travels along the z-direction its electric and magnetic field vectors are lying in the x-y plane as they are mutually perpendicular.

Frequency of wave = 30 MHz

Wavelength=

\frac{Speed\ of\ light}{Frequency}=\frac{3\times 10^{8}}{30\times 10^{6}}

=10m .

5. A radio can tune in to any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band?

Answer:

Frequency range = 7.5 MHz to 12 MHz

Speed of light = 3\times 10^{8}ms^{-1}

Wavelength corresponding to the frequency of 7.5 MHz =

\frac{3\times 10^{8}ms^{-1}}{7.5\times 10^{6}Hz}

=40m

Wavelength corresponding to the frequency of 12 MHz =

\frac{3\times 10^{8}ms^{-1}}{12\times 10^{6}Hz}

=25m

The corresponding wavelength band is 25m to 40m.

6. A charged particle oscillates about its mean equilibrium position with a frequency of 10^{9} Hz. What is the frequency of the electromagnetic waves produced by the oscillator?

Answer:

The frequency of the electromagnetic waves produced by the oscillation of a charged particle about a mean position is equal to the frequency of the oscillation of the charged particle. Therefore electromagnetic waves produced will have a frequency of 10 9 Hz.

7. The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is B_0 = 510 n T . What is the amplitude of the electric field part of the wave?

Answer:

Magnetic Field (B 0 )=510 nT = 510 \times 10 -9 T

Speed of light(c) = 3 \times 10 8 ms -1

Electric Field = B 0 \times c

= 510 \times 10 -9 T \times 3 \times 10 8 ms -1

= 153 NC -1

8. Suppose that the electric field amplitude of an electromagnetic wave is E_0 = 120 N/C and that its frequency is \nu =50.0\ MHz (a) Determine, B_0 , \omega , \ k , \lambda (b) Find expressions for E and B.

Answer:

E 0 = 120 NC -1

\nu =50.0\ MHz

(a)

Magnetic\ Field \ amplitude(B0) =\frac{E_{0}}{c}

= \frac{120}{3\times 10^{8}}

=400 nT

Angular frequency ( \omega ) = 2 \pi \nu

=2 \times \pi \times 50 \times 10^{6}

=3.14 \times 10 8 rad s -1

Propagation constant(k)

= \frac{2\pi }{\lambda }

= \frac{2\pi \nu }{\lambda\nu }

= \frac{\omega }{c}

= \frac{3.14\times 10^{8} }{3\times 10^{8}}

=1.05 rad m -1

Wavelength( \lambda ) = \frac{c}{\nu }

= \frac{3\times 10^{8}}{50\times 10^{6}} = 6 m

Assuming the electromagnetic wave propagates in the positive z-direction the Electric field vector will be in the positive x-direction and the magnetic field vector will be in the positive y-direction as they are mutually perpendicular and E_{0}\times B_{0} gives the direction of propagation of the wave.

\vec{E}=E_{0}sin(kx-\omega t)\hat{i}

= 120sin(1.05x-3.14\times 10^{8} t)NC^{-1}\hat{i}


\vec{B}=B_{0}sin(kx-\omega t)\hat{j}

= 400sin(1.05x-3.14\times 10^{8} t)\ nT\hat{j}

9) The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula E = hv (for energy of a quantum of radiation: photon) and obtain the photon energy in units of eV for different parts of the electromagnetic spectrum. In what way are the different scales of photon energies that you obtain related to the
sources of electromagnetic radiation?

Answer:

E=h\nu=\frac{hc}{\lambda}

E=h\nu=\frac{6.6\times 10^{-34}\times3\times10^8}{\lambda \times1.6\times10^{-19}}eV=\frac{12.375\times 10^{-7}}{\lambda}eV

Now substitute the different range of wavelength in the electromagnetic spectrum to obtain the energy

EM wave one wavelength is taken from the range Energy in eV
Radio 1 m 1.2375\times10^{-6}
Microwave 1 mm 1.2375\times10^{-3}
Infra-red 1000 nm 1.2375
Light 500 nm 2.475
Ultraviolet 1nm 1237.5
X-rays 0.01 nm 123750
Gamma rays 0.0001 nm 12375000

10.(c) In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 \times 10 ^{10} Hz and amplitude 48 V m^{-1} .

Show that the average energy density of the E field equals the average energy density of the B field. [c = 3 \times 10^8 m s^{-1}. ]

Answer:

The average energy density of the Electric field(U E )

= \frac{1}{2}\epsilon E^{2}

= \frac{1}{2}\epsilon (Bc)^{2} (as E=Bc)

= \frac{1}{2}\epsilon \frac{B^{2}}{\mu \epsilon } (c=\frac{1}{\sqrt{\mu \epsilon }})

=U B

Therefore the average energy density of the electric field is equal to the average energy density of the Magnetic field.

11 (a) Suppose that the electric field part of an electromagnetic wave in vacuum is E = \left \{ ( 3.1 N/C )\cos [ ( 1.8 rad /m)y + ( 5.4 \times 10 ^6 rad /s t )] \right \} \hat i
What is the direction of propagation?

Answer:

E = \left \{ ( 3.1 N/C )\cos [ ( 1.8 rad /m)y + ( 5.4 \times 10 ^6 rad /s t )] \right \} \hat i

E = \left \{ ( 3.1 N/C )\sin [ (\frac{\pi }{2}-(( 1.8 rad /m)y + ( 5.4 \times 10 ^6 rad /s t ))] \right \} \hat i

The electric field vector is in the negative x-direction and the wave propagates in the negative y-direction.

11 (b) Suppose that the electric field part of an electromagnetic wave in vacuum is

E = \left \{ ( 3.1 N/C )\cos [ ( 1.8 rad /m)y + ( 5.4 \times 10 ^6 rad /s t )] \right \} \hat i

What is the wavelength?

Answer:

From the equation of the wave given we can infer k = 1.8 rad m -1

Wavelength(\lambda )=\frac{2\pi }{k}

=\frac{2\pi }{1.8}

=3.49 m

11 (c) Suppose that the electric field part of an electromagnetic wave in vacuum is

E = \left \{ ( 3.1 N/C )\cos [ ( 1.8 rad /m)y + ( 5.4 \times 10 ^6 rad /s t )] \right \} \hat i

What is the frequency n?

Answer:

From the given equation of the electric field we can infer angular frequency( \omega ) = 5.4 \times 10 8 rad s -1

Frequency( \nu ) = \frac{\omega }{2\pi }

= \frac{5.4\times10^{8}}{2\pi }

=8.6 \times 10 7 Hz

=86 MHz

11 (d) Suppose that the electric field part of an electromagnetic wave in vacuum is

E = \left \{ ( 3.1 N/C )\cos [ ( 1.8 rad /m)y + ( 5.4 \times 10 ^6 rad /s t )] \right \} \hat i
What is the amplitude of the magnetic field part of the wave?

Answer:

From the given equation of the electric field, we can infer Electric field amplitude (E 0 ) =3.1 NC -1

Magnetic field amplitude (B_0) =\frac{E_{0}}{c}

=\frac{3.1}{3\times 10^{8}}

=1.03 \times 10 -7 T

11(e) Suppose that the electric field part of an electromagnetic wave in vacuum is

E = \left \{ ( 3.1 N/C )\cos [ ( 1.8 rad /m)y + ( 5.4 \times 10 ^6 rad /s )t] \right \} \hat i

Write an expression for the magnetic field part of the wave.

Answer:

As the electric field vector is directed along the negative x-direction and the electromagnetic wave propagates along the negative y-direction the magnetic field vector must be directed along the negative z-direction. ( -\hat{i}\times -\hat{k}=-\hat{j} )

Therefore, \vec{B}= \left \{ B_{0}\cos [ ( 1.8 rad /m)y + ( 5.4 \times 10 ^6 rad /s )t] \right \} \hat k

\vec{B}= \left \{ 1.03\times 10^{-7}\cos [ ( 1.8 rad /m)y + ( 5.4 \times 10 ^6 rad /s )t] \right \}T \hat k

12.(a) About 5% of the power of a 100 W light bulb is converted to visible radiation. What is the average intensity of visible radiation at a distance of 1m from the bulb?

Assume that the radiation is emitted isotropically and neglect reflection.

Answer:

Total power which is converted into visible radiation = 5% of 100W = 5W

The above means 5J of energy is passing through the surface of a concentric sphere(with the bulb at its centre) per second.

Intensity for a sphere of radius 1m

=\frac{5}{4\pi (1)^{2}}

=0.398 Wm -2

12.(b) About 5% of the power of a 100 W light bulb is converted to visible
radiation. What is the average intensity of visible radiation at a distance of 10 m?
Assume that the radiation is emitted isotropically and neglect
reflection.

Answer:

Total power which is converted into visible radiation = 5% of 100W = 5W

The above means 5J of energy is passing through the surface of a concentric sphere(with the bulb at its centre) per second.

Intensity for a sphere of radius 10 m

=\frac{5}{4\pi (10)^{2}}

=3.98 \times 10 -3 Wm -2

13) Use the formula \lambda _m T = 0.29 cm K to obtain the characteristic
temperature ranges for different parts of the electromagnetic
spectrum. What do the numbers that you obtain tell you?

Answer:

EM wave one wavelength is taken from the range Temperature T=\frac{0.29}{\lambda}
Radio 100 cm 2.9\times 10^{-3}K
Microwave 0.1cm 2.9 K
Infra-red 100000ncm 2900K
Light 50000 ncm 5800K
Ultraviolet 100ncm 2.9\times10^{6}K
X-rays 1 ncm 2.9\times10^8K
Gamma rays 0.01 nm 2.9\times10^{10}K


These numbers indicate the temperature ranges required for obtaining radiations in different parts of the spectrum

Answer the following questions

15. (a) Long distance radio broadcasts use short-wave bands. Why?

Answer:

Long-distance radio broadcasts use short-wave bands as these are refracted by the ionosphere.

Answer the following questions

15. (b) It is necessary to use satellites for long distance TV transmission. Why?

Answer:

As TV signals are of high frequencies they are not reflected by the ionosphere and therefore satellites are to be used to reflect them.

Answer the following questions

15 (c) Optical and radio telescopes are built on the ground but X-ray astronomy is possible only from satellites orbiting the earth. Why?

Answer:

X-rays are absorbed by the atmosphere and therefore the source of X-rays must lie outside the atmosphere to carry out X-ray astronomy and therefore satellites orbiting the earth are necessary but radio waves and visible light can penetrate through the atmosphere and therefore optical and radio telescopes can be built on the ground.

Answer the following questions

15. (d) The small ozone layer on top of the stratosphere is crucial for human survival. Why?

Answer:

The small ozone layer on top of the stratosphere is crucial for human survival as it absorbs the ultraviolet radiations coming from the sun which are very harmful to humans.

Answer the following questions

15. (e) If the earth did not have an atmosphere, would its average surface temperature be higher or lower than what it is now?

Answer:

If the earth did not have an atmosphere its average surface temperature be lower than what it is now as in the absence of atmosphere there would be no greenhouse effect.

Answer the following questions

15 f) Some scientists have predicted that a global nuclear war on the earth would be followed by a severe ‘nuclear winter’ with a devastating effect on life on earth. What might be the basis of this prediction?

Answer:

The use of nuclear weapons would cause the formation of smoke clouds preventing the light from the sun reaching earth surface and it will also deplete the atmosphere and therefore stopping the greenhouse effect and thus doubling the cooling effect.

Class 12 physics ch 8 NCERT solutions is an essential resource for students preparing for both board exams and competitive exams like JEE. This chapter can be considered moderately challenging but also offers scoring potential. These solutions provide a clear understanding of electromagnetic waves, making complex concepts manageable, and play a crucial role in helping students excel in both board exams and JEE, contributing significantly to their success. There are 15 questions in the Electromagnetic Waves NCERT solutions. The NCERT solution for Class 12 Physics chapter 8 covers theoretical as well as numerical problems.

NCERT solutions for class 12 physics chapter-wise

NCERT Solutions for Class 12 Physics Chapter 8: Important Formulas and Diagrams

  • Gauss’ law of electricity:

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{"code":"$$\\oint_{}^{}\\vec{E.}d\\vec{s}\\,=\\,\\frac{q}{\\epsilon_{0}}$$","backgroundColor":"#ffffff","backgroundColorModified":null,"font":{"color":"#000000","family":"Arial","size":11},"id":"3","aid":null,"type":"$$","ts":1637054812868,"cs":"J5PxCQICONS3+SPn+w+DFQ==","size":{"width":104,"height":37}}

  • Gauss’s law of magnetism:

{"aid":null,"id":"4","backgroundColorModified":false,"font":{"color":"#000000","size":11,"family":"Arial"},"backgroundColor":"#ffffff","type":"$$","code":"$$\\oint_{}^{}\\vec{B}.d\\vec{S}\\,=\\,0$$","ts":1637055016683,"cs":"15C3V/oS4nBin7YLhgWUDg==","size":{"width":97,"height":37}}

  • Faraday’s laws of Electromagnetic induction:

{"id":"5","aid":null,"font":{"color":"#000000","family":"Arial","size":11},"type":"$$","backgroundColor":"#ffffff","backgroundColorModified":false,"code":"$$e\\,=\\,-\\,\\frac{d\\phi_{B}}{dt}$$","ts":1637055464878,"cs":"numdiV7y2ABV1sdwjXgcNQ==","size":{"width":93,"height":34}}

Since emf can be defined as the line integral of the electric field, the above relation can be expressed as

{"id":"6","font":{"family":"Arial","color":"#000000","size":11},"type":"$$","backgroundColor":"#ffffff","aid":null,"backgroundColorModified":false,"code":"$$\\oint_{}^{}\\vec{E}.d\\vec{l}\\,=\\,-\\,\\frac{d\\phi_{B}}{dt}$$","ts":1637055605991,"cs":"2It52s47V8EVzWxmp+TIRw==","size":{"width":146,"height":37}}

  • Ampere-Maxwell’s Circuital law:

{"code":"$$\\oint_{}^{}\\vec{B.}d\\vec{l}\\,=\\,\\mu_{0}\\left(i\\,+\\,i_{d}\\right)$$","aid":null,"backgroundColor":"#ffffff","id":"7","type":"$$","backgroundColorModified":false,"font":{"size":11,"family":"Arial","color":"#000000"},"ts":1637056068499,"cs":"OGGYEsZmvZcrjuCOSYuwfg==","size":{"width":160,"height":37}}

Where

{"font":{"size":11,"color":"#000000","family":"Arial"},"backgroundColor":"#ffffff","backgroundColorModified":false,"code":"$$i_{d}\\,=\\,\\epsilon_{0}A\\frac{dE}{dt}$$","type":"$$","aid":null,"id":"8","ts":1637056118397,"cs":"UTVFqDR1iuL9gpEFg4Xrxg==","size":{"width":96,"height":34}}

Electromagnetic Spectrum

  • Gamma Rays:

Wavelength range: 1 x 10¹⁴ to 1 x 10⁻¹⁰m

Frequency range: 3 x 10²² to 3 x 10¹⁸Hz

Production: By transition of atomic nuclei and decay of certain elementary particles.

  • X- rays:

Wavelength range: 1 x 10⁻¹¹ to 3 x 10⁻⁸m

Frequency range: 3 x 10¹⁹ to 1 x 10¹⁶Hz

Production: By sudden deceleration of high-speed electrons at a high-atomic number target, and also by the electronic transition among the innermost orbits of atoms.

  • Ultraviolet- Rays:

Wavelength range: 1 x 10⁻⁸ to 3 x 10⁻⁸m

Frequency range: 3 x 10¹⁶ to 1 x 10¹⁴Hz

Production: By sun, arc, vacuum spark, and ionized gases.

  • Visible Light:

Wavelength range: 1 x 10⁻⁷ to 1 x 10⁻⁷m

Frequency range: 7 x 10¹⁴ to 4 x 10¹⁴Hz

Production: Radiated by excited atoms in gases and incandescent bodies.

  • Infrared Radiation

Wavelength range: 8 x 10⁻⁷ to 5 x 10⁻³m

Frequency range: 4x 10¹⁴ to 6 x 10¹⁰ Hz

Production: From hot bodies, and by rotational and vibrational transitions in molecules

  • Radio waves

Wavelength range: 1 x 10⁻¹ to 1 x 10⁴m

Frequency range: 3 x 10⁹ to 3 x 10⁴ Hz

Production: By oscillating electric circuits.

Also, check

Electromagnetic Wave Class 12 Physics NCERT: Topics

The following are the main headings covered in the Chapter 8 Physics Class 12-

  • Concepts of displacement current and Maxwell's equation
  • Basics of electromagnetic waves
  • Electromagnetic Spectrum

Significance of NCERT solutions for class 12 physics chapter 8 electromagnetic induction:

  • For the CBSE board exam, 4% of questions are expected from chapter 8 physics class 12.

  • NCERT solutions for class 12 will give a clear idea about how to use the formulas studied in the NCERT chapter 8 Physics Class 12.

  • For exams like NEET and JEE Main one or two questions are asked from the Electromagnetic Waves NCERT.

  • Questions based on the wave equation are expected from the chapter for the CBSE board exam. The relation between electric field, magnetic field and speed of light is also important.

  • The electromagnetic Waves Class 12 pdf download button is available for the ease of students to prepare offline using Electromagnetic Waves Class 12 NCERT solutions.

Key Features of Electromagnetic Waves Class 12 NCERT SolutionsPDF

  1. Comprehensive Coverage: These ncert solutions electromagnetic waves comprehensively cover all the topics and questions presented in the Class 12 Physics chapter on electromagnetic waves.

  2. Detailed Explanations: Each class 12 chapter 8 physics ncert solutions provides detailed step-by-step explanations, simplifying complex concepts for students.

  3. Clarity and Simplicity: The solutions are presented in clear and simple language, ensuring ease of understanding.

  4. Exam Preparation: These solutions are crucial for board exam preparation and provide valuable support for competitive exams like JEE.

  5. Balanced Complexity: The chapter balances complexity with scoring potential, making it accessible for students aiming to score well.

  6. Free Access: These solutions are available for free, ensuring equal access for all students.

Excluded Content:

Certain portions have been omitted, which include:

  • Example 8.1.
  • Section 8.3.2 discusses the nature of electromagnetic waves (excluding references to ether and content on page 277).
  • Examples 8.4 and 8.5.
  • Exercises 8.11 to 8.15.

Also Check NCERT Books and NCERT Syllabus here:

NCERT solutions subject wise

Frequently Asked Question (FAQs)

1. What is the prerequisite of the NCERT chapter electromagnetic waves?

Before starting the preparation of NCERT class 12 chapter 8 it is better to have an idea of chapters 1 to 7. Chapter 1 to 7 is interconnected.

2. How many questions are expected from Class 12 Physics Chapter 8 NCERT solutions for NEET?

One question from chapter 8 electromagnetic waves can be expected for the NEET exam. The questions can be theoretical or numerical.

3. What is the weightage of the NCERT class 12 chapter electromagnetic wave for JEE Main?

From the chapter on electromagnetic waves, one question can be expected for JEE Main. More than one question can also be asked.

4. How many marks are allotted to electromagnetic wave for CBSE board exams?

3 to 5 marks questions are asked from class 12 chapter electromagnetic waves for CBSE board exams.

5. How can students complete electromagnetic waves ncert solutions for chapter 8 physics class 12 with a perfect score?

Experts at Careers360 have created the chapter 8 physics class 12 ncert solutions after completing extensive research on each concept. To assist students perform well on class assignments and board exams, every small element is well addressed. Additionally, it enables students to successfully complete their tasks on time.

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

hello,

Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.

I hope this was helpful!

Good Luck

Hello dear,

If you was not able to clear 1st compartment and now you giving second compartment so YES, you can go for your improvement exam next year but if a student receives an improvement, they are given the opportunity to retake the boards as a private candidate the following year, but there are some requirements. First, the student must pass all of their subjects; if they received a compartment in any subject, they must then pass the compartment exam before being eligible for the improvement.


As you can registered yourself as private candidate for giving your improvement exam of 12 standard CBSE(Central Board of Secondary Education).For that you have to wait for a whole year which is bit difficult for you.


Positive side of waiting for whole year is you have a whole year to preparing yourself for your examination. You have no distraction or something which may causes your failure in the exams. In whole year you have to stay focused on your 12 standard examination for doing well in it. By this you get a highest marks as a comparison of others.


Believe in Yourself! You can make anything happen


All the very best.

Hello Student,

I appreciate your Interest in education. See the improvement is not restricted to one subject or multiple subjects  and  we cannot say if improvement in one subject in one year leads to improvement in more subjects in coming year.

You just need to have a revision of all subjects what you have completed in the school. have a revision and practice of subjects and concepts helps you better.

All the best.

If you'll do hard work then by hard work of 6 months you can achieve your goal but you have to start studying for it dont waste your time its a very important year so please dont waste it otherwise you'll regret.

Yes, you can take admission in class 12th privately there are many colleges in which you can give 12th privately.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

Bio Medical Engineer

The field of biomedical engineering opens up a universe of expert chances. An Individual in the biomedical engineering career path work in the field of engineering as well as medicine, in order to find out solutions to common problems of the two fields. The biomedical engineering job opportunities are to collaborate with doctors and researchers to develop medical systems, equipment, or devices that can solve clinical problems. Here we will be discussing jobs after biomedical engineering, how to get a job in biomedical engineering, biomedical engineering scope, and salary. 

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Data Administrator

Database professionals use software to store and organise data such as financial information, and customer shipping records. Individuals who opt for a career as data administrators ensure that data is available for users and secured from unauthorised sales. DB administrators may work in various types of industries. It may involve computer systems design, service firms, insurance companies, banks and hospitals.

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Ethical Hacker

A career as ethical hacker involves various challenges and provides lucrative opportunities in the digital era where every giant business and startup owns its cyberspace on the world wide web. Individuals in the ethical hacker career path try to find the vulnerabilities in the cyber system to get its authority. If he or she succeeds in it then he or she gets its illegal authority. Individuals in the ethical hacker career path then steal information or delete the file that could affect the business, functioning, or services of the organization.

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Data Analyst

The invention of the database has given fresh breath to the people involved in the data analytics career path. Analysis refers to splitting up a whole into its individual components for individual analysis. Data analysis is a method through which raw data are processed and transformed into information that would be beneficial for user strategic thinking.

Data are collected and examined to respond to questions, evaluate hypotheses or contradict theories. It is a tool for analyzing, transforming, modeling, and arranging data with useful knowledge, to assist in decision-making and methods, encompassing various strategies, and is used in different fields of business, research, and social science.

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Geothermal Engineer

Individuals who opt for a career as geothermal engineers are the professionals involved in the processing of geothermal energy. The responsibilities of geothermal engineers may vary depending on the workplace location. Those who work in fields design facilities to process and distribute geothermal energy. They oversee the functioning of machinery used in the field.

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Remote Sensing Technician

Individuals who opt for a career as a remote sensing technician possess unique personalities. Remote sensing analysts seem to be rational human beings, they are strong, independent, persistent, sincere, realistic and resourceful. Some of them are analytical as well, which means they are intelligent, introspective and inquisitive. 

Remote sensing scientists use remote sensing technology to support scientists in fields such as community planning, flight planning or the management of natural resources. Analysing data collected from aircraft, satellites or ground-based platforms using statistical analysis software, image analysis software or Geographic Information Systems (GIS) is a significant part of their work. Do you want to learn how to become remote sensing technician? There's no need to be concerned; we've devised a simple remote sensing technician career path for you. Scroll through the pages and read.

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Geotechnical engineer

The role of geotechnical engineer starts with reviewing the projects needed to define the required material properties. The work responsibilities are followed by a site investigation of rock, soil, fault distribution and bedrock properties on and below an area of interest. The investigation is aimed to improve the ground engineering design and determine their engineering properties that include how they will interact with, on or in a proposed construction. 

The role of geotechnical engineer in mining includes designing and determining the type of foundations, earthworks, and or pavement subgrades required for the intended man-made structures to be made. Geotechnical engineering jobs are involved in earthen and concrete dam construction projects, working under a range of normal and extreme loading conditions. 

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Cartographer

How fascinating it is to represent the whole world on just a piece of paper or a sphere. With the help of maps, we are able to represent the real world on a much smaller scale. Individuals who opt for a career as a cartographer are those who make maps. But, cartography is not just limited to maps, it is about a mixture of art, science, and technology. As a cartographer, not only you will create maps but use various geodetic surveys and remote sensing systems to measure, analyse, and create different maps for political, cultural or educational purposes.

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Budget Analyst

Budget analysis, in a nutshell, entails thoroughly analyzing the details of a financial budget. The budget analysis aims to better understand and manage revenue. Budget analysts assist in the achievement of financial targets, the preservation of profitability, and the pursuit of long-term growth for a business. Budget analysts generally have a bachelor's degree in accounting, finance, economics, or a closely related field. Knowledge of Financial Management is of prime importance in this career.

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Data Analyst

The invention of the database has given fresh breath to the people involved in the data analytics career path. Analysis refers to splitting up a whole into its individual components for individual analysis. Data analysis is a method through which raw data are processed and transformed into information that would be beneficial for user strategic thinking.

Data are collected and examined to respond to questions, evaluate hypotheses or contradict theories. It is a tool for analyzing, transforming, modeling, and arranging data with useful knowledge, to assist in decision-making and methods, encompassing various strategies, and is used in different fields of business, research, and social science.

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Product Manager

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Underwriter

An underwriter is a person who assesses and evaluates the risk of insurance in his or her field like mortgage, loan, health policy, investment, and so on and so forth. The underwriter career path does involve risks as analysing the risks means finding out if there is a way for the insurance underwriter jobs to recover the money from its clients. If the risk turns out to be too much for the company then in the future it is an underwriter who will be held accountable for it. Therefore, one must carry out his or her job with a lot of attention and diligence.

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Finance Executive
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Individuals in the operations manager jobs are responsible for ensuring the efficiency of each department to acquire its optimal goal. They plan the use of resources and distribution of materials. The operations manager's job description includes managing budgets, negotiating contracts, and performing administrative tasks.

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Investment Director

An investment director is a person who helps corporations and individuals manage their finances. They can help them develop a strategy to achieve their goals, including paying off debts and investing in the future. In addition, he or she can help individuals make informed decisions.

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Welding Engineer

Welding Engineer Job Description: A Welding Engineer work involves managing welding projects and supervising welding teams. He or she is responsible for reviewing welding procedures, processes and documentation. A career as Welding Engineer involves conducting failure analyses and causes on welding issues. 

5 Jobs Available
Transportation Planner

A career as Transportation Planner requires technical application of science and technology in engineering, particularly the concepts, equipment and technologies involved in the production of products and services. In fields like land use, infrastructure review, ecological standards and street design, he or she considers issues of health, environment and performance. A Transportation Planner assigns resources for implementing and designing programmes. He or she is responsible for assessing needs, preparing plans and forecasts and compliance with regulations.

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Plumber

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Construction Manager

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Urban Planner

Urban Planning careers revolve around the idea of developing a plan to use the land optimally, without affecting the environment. Urban planning jobs are offered to those candidates who are skilled in making the right use of land to distribute the growing population, to create various communities. 

Urban planning careers come with the opportunity to make changes to the existing cities and towns. They identify various community needs and make short and long-term plans accordingly.

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Highway Engineer

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Environmental Engineer

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Naval Architect

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Orthotist and Prosthetist

Orthotists and Prosthetists are professionals who provide aid to patients with disabilities. They fix them to artificial limbs (prosthetics) and help them to regain stability. There are times when people lose their limbs in an accident. In some other occasions, they are born without a limb or orthopaedic impairment. Orthotists and prosthetists play a crucial role in their lives with fixing them to assistive devices and provide mobility.

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Speech Therapist
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Audiologist

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For an individual who opts for a career as an actor, the primary responsibility is to completely speak to the character he or she is playing and to persuade the crowd that the character is genuine by connecting with them and bringing them into the story. This applies to significant roles and littler parts, as all roles join to make an effective creation. Here in this article, we will discuss how to become an actor in India, actor exams, actor salary in India, and actor jobs. 

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Video Game Designer

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Depending on the video game designer job description and experience they may also have to lead a team and do the early testing of the game in order to suggest changes and find loopholes.

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Radio Jockey

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A career as radio jockey has a lot to offer to deserving candidates. If you want to know more about a career as radio jockey, and how to become a radio jockey then continue reading the article.

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Choreographer

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Videographer
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Multimedia Specialist

A multimedia specialist is a media professional who creates, audio, videos, graphic image files, computer animations for multimedia applications. He or she is responsible for planning, producing, and maintaining websites and applications. 

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Social Media Manager

A career as social media manager involves implementing the company’s or brand’s marketing plan across all social media channels. Social media managers help in building or improving a brand’s or a company’s website traffic, build brand awareness, create and implement marketing and brand strategy. Social media managers are key to important social communication as well.

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Copy Writer

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Journalist

Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.

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Publisher

For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.

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Vlogger

In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion. 

Ever since internet costs got reduced the viewership for these types of content has increased on a large scale. Therefore, a career as a vlogger has a lot to offer. If you want to know more about the Vlogger eligibility, roles and responsibilities then continue reading the article. 

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Editor

Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.

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Linguist

Linguistic meaning is related to language or Linguistics which is the study of languages. A career as a linguistic meaning, a profession that is based on the scientific study of language, and it's a very broad field with many specialities. Famous linguists work in academia, researching and teaching different areas of language, such as phonetics (sounds), syntax (word order) and semantics (meaning). 

Other researchers focus on specialities like computational linguistics, which seeks to better match human and computer language capacities, or applied linguistics, which is concerned with improving language education. Still, others work as language experts for the government, advertising companies, dictionary publishers and various other private enterprises. Some might work from home as freelance linguists. Philologist, phonologist, and dialectician are some of Linguist synonym. Linguists can study French, German, Italian

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Public Relation Executive
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Travel Journalist

The career of a travel journalist is full of passion, excitement and responsibility. Journalism as a career could be challenging at times, but if you're someone who has been genuinely enthusiastic about all this, then it is the best decision for you. Travel journalism jobs are all about insightful, artfully written, informative narratives designed to cover the travel industry. Travel Journalist is someone who explores, gathers and presents information as a news article.

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Welding Engineer

Welding Engineer Job Description: A Welding Engineer work involves managing welding projects and supervising welding teams. He or she is responsible for reviewing welding procedures, processes and documentation. A career as Welding Engineer involves conducting failure analyses and causes on welding issues. 

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QA Manager
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Quality Controller

A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product. 

A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

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Product Manager

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QA Lead

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Metallurgical Engineer

A metallurgical engineer is a professional who studies and produces materials that bring power to our world. He or she extracts metals from ores and rocks and transforms them into alloys, high-purity metals and other materials used in developing infrastructure, transportation and healthcare equipment. 

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Azure Administrator

An Azure Administrator is a professional responsible for implementing, monitoring, and maintaining Azure Solutions. He or she manages cloud infrastructure service instances and various cloud servers as well as sets up public and private cloud systems. 

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AWS Solution Architect

An AWS Solution Architect is someone who specializes in developing and implementing cloud computing systems. He or she has a good understanding of the various aspects of cloud computing and can confidently deploy and manage their systems. He or she troubleshoots the issues and evaluates the risk from the third party. 

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QA Manager
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Computer Programmer

Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

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A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

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Information Security Manager

Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack 

3 Jobs Available
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