Question

Figure shows a conducting loop consisting of a half-circle of radius $r=0.20m$ and three straight sections. The half-circle lies in a uniform magnetic field $B$ that is directed out of the page, the field magnitude is given by $B=(4.0T/s^2)t^2+(2.0T/s)t+3.0T$ An ideal battery with ${\epsilon }_0=2.0V$ is connected to the loop. The resistance of the loop is $2.0\Omega$ . The current in the loop at $t=10s$ will be close to

• A. 3.6 A
• B. 1.6 A
• C. 6.2 A
• D. 4.2 A

Answer 1

Answer: (C)

Given that $r=0.20m$,
$t=10s$,
$R=2\Omega$
$B=\left(4.0T/S^2\right)t^2+(2.0T/S)t+3.0T$
$\frac{dB}{dt}=8t+2$
From $E=-\frac{d\varphi }{dt}=-A\frac{d\varphi }{dt}$
$E=-\pi r^2[8t+2]$
$=3.14\times 0.2\times 0.2\times 82=10.30$
Total $E=10.30+2$
$=12.30V$
$I=\frac{E}{R}=\frac{12.30}{2.0}$
$=6.15\approx 6.2A$