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  2. Solve it, - Electromagnetic Induction and Alternating currents - JEE Main-4
# Engineering

The figure shows a square loop L of side 5 cm which is connected to a network of resistances. The whole setup is moving towards right with a constant speed of 1 cm s^{-1}. At some instant, a part of L is in a uniform magnetic field of 1 T, perpendicular to the plane of the loop. If the resistance of L is 1.7\Omega, the current in the loop at that instant will be close to :

  • Option 1)

    150 \mu A

  • Option 2)

    115 \mu A

  • Option 3)

    60 \mu A

  • Option 4)

    170 \mu A

 

Answers (1)
S solutionqc

 

Motional EMF -

\varepsilon = Blv
 

- wherein

B\rightarrow magnetic field

l\rightarrow length

u\rightarrow velocity of u perpendicular to uniform magnetic field.

 

 

B=1T

L=5\times10^{-2}m

v=1\times10^{-2}m/s

E=BVL=5\times10^{-4}V

The circuit ABCD is a wheat stone bridge circuit 

\Rightarrow Req=\frac{4\times2}{4+2}=\frac{4}{3}\Omega

It is given the resistance of square = 1.7\Omega

R total =\frac{4}{3}+1.7\Omega

I=\frac{V}{R}=\frac{5\times10^{-4}}{\frac{4}{3}+1.7}=170 \mu A


Option 1)

150 \mu A

Option 2)

115 \mu A

Option 3)

60 \mu A

Option 4)

170 \mu A

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