Question

A toroid having a rectangular cross-section $(a=2.00\text{cm}$ and $b=3.00\text{cm})$ and inner radius $r=4.00\text{cm}$ consists of $500$ turns of wire that carries a current $I=I_{max}\sin \omega t$ with $I_{max}=50.0\text{A}$ and a frequency $f=\frac{\omega }{2\pi }=60.0\text{Hz}.$ A coil that consists of $20$ turns of wire links with the toroid as shown in figure. Then which of the following is/are correct?

• A.
• B. Maximum value of induced emf is $0.422V$
• C.
• D. Maximum value of induced emf is $0.122V$

Answer 1

Answer: (B), (C)

Field due to the toroid is,

$B=\frac{{\mu }_{o}NI}{2\pi r}=\frac{500{\mu }_{o}I}{2\pi r}$

Flux linked with the coil is,

${\varphi }_B=\int BdA=\frac{500{\mu }_0{I}_{max}\sin \omega t}{2\pi }{\int }_R^{b+R}\frac{adr}{r}$

${\varphi }_B=\frac{500{\mu }_0{I}_{max}}{2\pi }a\sin \omega t\ln \left(\frac{b+R}{R}\right)$
$E=\frac{Nd{\varphi }_B}{dt}=20\left(\frac{500{\mu }_0{I}_{max}}{2\pi }\right)\omega a\ln \left(\frac{b+R}{R}\right)\cos \omega t$
$E=\frac{{10}^4}{2\pi }(4\pi \times {10}^{-7})\left(50\right)\left(377\right)\left(0.020\right)\ln \left(\frac{3+4}{4}\right)\cos \omega t$
$E=\left(0.422\right)\cos \omega t$

For maximum value of induced emf $\cos \omega t=1$

$E=0.422\text{V}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right),\&\left(C\right)$ is the correct answer.