JEE MAIN - Physics (2021 - 16th March Evening Shift - No. 14)
The magnetic field in a region is given by $$\overrightarrow B = {B_o}\left( {{x \over a}} \right)\widehat k$$. A square loop of side d is placed with its edges along the x and y axes. The loop is moved with a constant velocity $$\overrightarrow v $$ = v0$$\widehat i$$. The emf induced in the loop is :
_16th_March_Evening_Shift_en_14_1.png)
$${{{B_o}{v_o}{d^2}} \over {2a}}$$
$${{{B_o}v_o^2d} \over {2a}}$$
$${{{B_o}{v_o}d} \over {2a}}$$
$${{{B_o}{v_o}{d^2}} \over a}$$
Explanation
_16th_March_Evening_Shift_en_14_2.png)
We know, Emf = Blv
where B, l and v are mutually perpendicular to each other. If any two are not perpendicular to each other then Emf = 0.
Here for part AB and DC, length l and velocity v is parallel to each other so, Emf by AB and DC is zero.
Given $$\overrightarrow B = {B_0}\left( {{x \over a}} \right)\widehat k$$
At x = 0,
$$B = {B_0}\left( {{0 \over a}} \right)$$
$$ = 0$$
$$ \therefore $$ For part AD, Emf = 0
In BC part,
x = d
$$ \therefore $$ $$B = {B_0}\left( {{d \over a}} \right)$$
And Emf for BC part,
$$\varepsilon $$ $$ = Bvd$$
Here B, v, d are mutually perpendicular to each other.
$$ \therefore $$ $$\varepsilon $$ $$ = {B_0}\left( {{d \over a}} \right) \times {v_0} \times d$$
$$ = {{{B_0}{v_0}{d^2}} \over a}$$
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