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  • (a) A circular loop of radius carries a current . Obtain an expression for the magnetic field at a point on its axis

(a) A circular loop of radius R carries a current I. Obtain an expression for the magnetic field at a point on its axis at a distance
x from its centre.

(b) A conducting rod of length2\; m is placed on a horizontal table in north-south direction. It carries a current of 5\; A from south to
north. Find the direction and magnitude of the magnetic force acting on the rod. Given that the Earth’s magnetic field at the
place is 0.6\times 10^{-4}T and angle of dip is \frac{\pi }{6}

 

 

 

 
 
 
 
 

Answers (1)

a) 


Consider two small current elements on the circular loop which are opposite to each other.
then the magnetic field at a point p which is at a distance X from the centre of the coil is due to the x component of the fields of A and B and y components cancel out.
\because dB_{X}= \frac{\mu_{0}}{4\pi }\frac{Idl\times \sin \left ( 90 \right )}{\left ( \sqrt{R^{2}+X^{2}} \right )^{2}}\cos \theta
  dB_{X}= \frac{\mu_{0}Idl}{4\pi }\times \frac{1}{\left ( \sqrt{R^{2}+X}^{2} \right )^{2}}\times \frac{R}{\sqrt{R^{2}+X^{2}}}
            = \frac{\mu_{0}Idl}{4\pi }\frac{R}{\left ( X^{2}+R^{2} \right )^{\frac{3}{2}}}
B= \int dBX= \frac{\mu_{0}IR}{4\pi \left ( x^{2}+R^{2} \right )^{\frac{3}{2}}}\int dl
   = \frac{\mu_{0}I\times 2\pi R^{2}}{4\pi \left ( x^{2}+R^{2} \right )^{\frac{3}{2}}}
   = \frac{\mu_{0}IR^{2}}{2\left ( x^{2}+R^{2} \right )^{\frac{3}{2}}}

b)i)

\\F=ilB sin\theta=5\times2\times0.6\times10^{-4}sin\pi/6\\=5\times2\times0.6\times10^{-4}\times0.5\\=3\times10^{-4}N

ii) Towards east

Posted by

Safeer PP

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