Question 84: Electromagnetic Induction

Question

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 $$\overrightarrow v $$ = v
0
$$\widehat i$$. The emf induced in the loop is :

Accepted Answer

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 $$\overrightarrow B  = {B_0}\left( {{x \over a}} ight)\widehat k$$
At x = 0,
$$B = {B_0}\left( {{0 \over a}} ight)$$
$$ = 0$$
$$ 	herefore $$ For part AD, Emf = 0
In BC part,
x = d
$$ 	herefore $$ $$B = {B_0}\left( {{d \over a}} ight)$$
And Emf for BC part,
$$\varepsilon $$ $$  = Bvd$$
Here B, v, d are mutually perpendicular to each other.
$$ 	herefore $$ $$\varepsilon $$ $$  = {B_0}\left( {{d \over a}} ight) 	imes {v_0} 	imes d$$
$$ = {{{B_0}{v_0}{d^2}} \over a}$$

Answer

$${{{B_o}{v_o}{d^2}} \over {2a}}$$

Answer

$${{{B_o}v_o^2d} \over {2a}}$$

Answer

$${{{B_o}{v_o}d} \over {2a}}$$

Answer

$${{{B_o}{v_o}{d^2}} \over a}$$

Electromagnetic Induction

The magnetic field in a region is given by $\overrightarrow B = {B_o}\left( {{x \over a}} \right)\widehat k. A square loop of side d is placed with its edges along the x and y axes. The loop is moved with a constant velocity \overrightarrow v = v 0 \widehat i$. The emf induced in the loop is :

Choose the correct answer: