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  • A magnetic field <img alt="\mathrm{B}=\frac{\mathrm {B}_0 \mathrm {y}}{\mathrm {a}} \hat{\mathrm{k}}" src="/latex-image/?%5Cmathrm%7BB%7D%3D%5Cfrac%7B%5Cmathrm%20%7BB%7D_0%20%5Cmathrm%20%7By%7D%7D%

A magnetic field \mathrm{B}=\frac{\mathrm {B}_0 \mathrm {y}}{\mathrm {a}} \hat{\mathrm{k}}  is directed into the paper in the +Z direction. B0 , a are the positive constants. A square loop EFGH of side a mass m and resistance R, in xy
plane, starts falling under the influence of gravity the induced current in the loop and indicate its direction is

Option: 1

\frac{\mathrm{B}_0 \mathrm{av}}{\mathrm{R}} \: \, \mathrm {anticlockwise}


Option: 2

\frac{\mathrm{B}_0 \mathrm{av}}{\mathrm{R}} \: \: \mathrm{clockwise}


Option: 3

\frac{2 \mathrm{~B}_0 \mathrm{av}}{\mathrm{R}} \: \: \mathrm {anticlockwise}


Option: 4

\mathrm {\frac{2 B_0 \text { av }}{R}}\: \: \mathrm {clockwise}


Answers (1)

best_answer

Let v = velocity of the loop at any time t.
 The induced emf between E & F

=\mathrm {\rho_1=B_1 \ell v=\frac{B_0 y}{a} \ell v=B_0 y v}

The induced emf between H and G

\begin{aligned} &\mathrm { =\rho_2=B_2 \ell v} \\ & \mathrm {=\frac{B_0(y+a)}{a} \ell v=B_0(y+a) v} \end{aligned}

 the net emf =\mathrm {\rho=\rho_2-\rho_1=B_0 a v}
 The induced current =\mathrm{i}=\frac{\rho}{\mathrm{R}}=\frac{\mathrm{B}_0 \mathrm{av}}{\mathrm{R}}   anticlockwise because

\rho_2>\rho_1

Posted by

Sumit Saini

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