Question

A conductor $AB$ of length $l$ moves in x-y plane with velocity $\overrightarrow{v}=v_0(\hat{i}-\hat{j})$. A magnetic field $\overrightarrow{B}=B_0(\hat{i}+\hat{j})$ exists in the region. The induced emf is

• A. zero
• B. $2B_{0}l{v}_0$
• C. $B_{0}l{v}_0$
• D. $\sqrt{2}{B}_{0}l{v}_0$

Answer 1

The correct option is A zero

Given, $B=B0(+);V=V0(-)$

Since, $e=-d\varphi /dtand\varphi =B.A$

So, $e=-d(B.A)/dt$

$e=d(B.A)/dt$ (Negative sign is neglected as it only represents the direction)

$e=B.d\left(A\right)/dt$ (Magnetic field is constant)
$=B.d\left(lb\right)/dt$ (Where l = Length and b = Breadth)

$=lB.d\left(b\right)/dt$ ---- (1)

As, $db/dt$ is nothing but the velocity with which the rod is moving.

So, $\frac{db}{dt}=V$

Putting the above value of $db/dt$ in equation (1).

We have, $e=lB.V$
$=lB0(+).V0(-)$

$=lB0V0(+).(-)$

$=lB0V0(1-1)$
$e=0$