Browse by Stream
QnA
Home

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • A conducting coil has 100 turns of area 2 m2 is placed in

A conducting coil has 100 turns of area 2 m2 is placed in a constant magnetic field of strength 0.5 mT . The coil is revolving with frequency 50 cycles / sec. The max current induced in coil when it has a resistance of 25\ohm\Omega .

Option: 1

2.4 A


Option: 2

2.31 A


Option: 3

1.25 A


Option: 4

2.8 A


Answers (1)

best_answer

As we learnt ,

 

Current -

i= \frac{\varepsilon }{R}= \frac{\varepsilon_{0}\sin \omega t }{R}= {i}'_{0\sin \omega t}

 

- wherein

\frac{\varepsilon_{0} }{R}=i{}'_{0}

 

 induced current , i = \frac{\varepsilon }{R}

i = \frac{NBA \omega sin \omega t}{R}

For max induced current \left | sin \omega t \right | = 1

i_{max} = \frac{NBA \omega }{R}

i_{max} = \frac{NBA 2 \pi f }{R}

i_{max} = \frac{100 \times 0.5 \times 10^{-3} \times 2 \times 2 \times \pi \times 50 }{25}

i_{max} = 1.25 A

 

Posted by

Rishabh

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

'Why are they dying only in Kota': SC grills Rajasthan over surge in student suicides
anam-khan

JEE Main 2025 paper 2 final answer key out; NTA drops 4 questions, 2 from BPlanning
anam-khan

JEE Main 2025 session 2 answer key out for BArch, BPlanning; response sheet on jeemain.nta.nic.in

Latest Question