Question

Figure shows a conducting loop consisting of a half-circle of radius $r = 0.20\,m$ 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=(4.0\,T/s^{2})t^{2}+(2.0\,T/s)t+3.0\,T $ An ideal battery with $ \varepsilon_{0}=2.0\,V $ is connected to the loop. The resistance of the loop is $ 2.0\,\Omega $ . The current in the loop at $ t=10\,s $ 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 $r =0.20\, m$,
$t =10 \,s$,
$R=2\, \Omega$
$B=\left(4.0\, T / S^{2}\right) t^{2}+(2.0 \,T / S) t+3.0\, T$
$ \frac{d B}{d t}=8 t+2$
From $E=-\frac{d \phi}{d t}=-A \frac{d \phi}{d t}$
$ E=-\pi r^{2}[8 t+2]$
$=3.14 \times 0.2 \times 0.2 \times 82=10.30$
Total $E=10.30+2$
$=12.30 \,V $
$I=\frac{E}{R}=\frac{12.30}{2.0}$
$=6.15 \approx 6.2 \,A$