Question

A square loop of side $1m$ is placed in a perpendicular magnetic field. Half of the area of the loop inside the magnetic field. A battery of emf $10V$ and negligible internal resistance is connected in the loop. The magnetic field changes with time according to relation $B=\left(0 . 01 - 2 t\right)Tesla$ . The resultant emf in the loop will be

• A. $1V$
• B. $11V$
• C. $10V$
• D. $9V$

Answer 1

Answer: (D)

Given $B=0.01-2tTesla$ ;
$\frac{d B}{d t}=-2Teslasec^{- 1}$ ,
Induced $emf$ $e=-\frac{d \phi}{d t}=-\frac{d}{d t}\left(B A\right)$
$=-A\frac{d B}{d t}=-\frac{1}{2}\left(1^{2}\right)\times \left(- 2\right)$
$\Rightarrow e=1V$
Since magnetic field $\left(\times \right)$ decreasing so according to Lenz’s law direction of induced current in upper part of square will be clockwise i.e. from $A$ to $C$ or in other words emf induces in a direction opposite to the main emf so resultant $emf=10-1=9V$ .