Question

The figure shows a conducting loop consisting of a half circle of area $A=0.06m^2$ and three straight segments.
The half circle lies in a uniform changing magnetic field $B=4t^2+2t+5$ (SI unit), where t is the time in
seconds.
An ideal battery E = 2 V is connected as shown and the total resistance of the wire is $2\Omega$. The net current in the loop at t = 5 seconds is:

• A. 1 A
• B. 1.5 A
• C. 0.26 A
• D. 0.10 A

Answer 1

The correct option is C 0.26 A

$|e_{ind}|=\frac{d\varphi }{dt}=\frac{d}{dt}\left(BA\right)=A\frac{dB}{dt}$

$=A\frac{d}{dt}(4t^2+2t+5)=A(8t+2)$

$=0.06(8t+2)$
which will come out to be 2.52 at t = 5.

$So,i=\frac{|2-e_{ind}|}{2}=0.26A$