Question

A copper wire having resistance of $0.01\Omega$ is used to wind a $400$ turns toroid of inner radius of $2\text{cm}$ and outer radius of $3\text{cm}$. Find the emf of a battery which when connected across the toroid will create a magnetic field of $1.0\times {10}^{-2}\text{T}$ at a distance equal to the mean radius of the toroid, from the center of the toroid.

• A. $31.25\text{mV}$
• B. $30\text{mV}$
• C. $25\text{mV}$
• D. $21.25\text{mV}$

Answer 1

The correct option is A $31.25\text{mV}$

Given:
$R=0.01\Omega ;N=400;B={10}^{-2}\text{T}$
$r_1=2\text{cm};r_2=3\text{cm}$

$r_{mean}=\frac{r_1+r_2}{2}=2.5\text{cm}=2.5\times {10}^{-2}\text{m}$

Let the emf of the battery be $E$.

Current flowing through the coil is,

$I=\frac{E}{R}=\frac{E}{0.01}=100E$

The magnetic field inside the toroid is,

$B=\frac{{\mu }_{0}NI}{2\pi r}$

${10}^{-2}=\frac{4\pi \times {10}^{-7}\times 400\times 100E}{2\pi \times 2.5\times {10}^{-2}}$

$E=\frac{2.5\times {10}^{-2}\times {10}^{-2}}{2\times {10}^{-7}\times 4\times {10}^4}=3.125\times {10}^{-2}\text{V}$

$\therefore E=31.25\text{mV}$

Hence, option $\left(A\right)$ is the correct answer.