Question

A toroid of $N$ turns, mean radius $r$ and cross- sectional radius $a$ is connected across a battery of emf $V$. If the magnetic field inside the core of toroid is $B$, find the value of $V$. Take resistance per unit length of wire equal to $\rho$.
[wires are closely wounded on toroid]

• A. $\frac{2{\pi }^{2}Nra\rho B}{{\mu }_{0}N}$
• B. $\frac{2{\pi }^{3}Nra\rho B}{{\mu }_{0}N}$
• C. $\frac{2\pi Nra\rho B}{{\mu }_{0}N}$
• D. $\frac{4{\pi }^{2}Nra\rho B}{{\mu }_{0}N}$

Answer 1

The correct option is D $\frac{4{\pi }^{2}Nra\rho B}{{\mu }_{0}N}$

Length of one term, $l=2\pi a$

Since wires are closely wounded and total number of turn is $N$.

Total length of wire in toroid will be

$L=Nl=2\pi Na$

So its total resistance $R$ will be

$R=L\times$ resistance per unit length $\left(\rho \right)$

$R=L\rho =2\pi Na\rho$

So current passes throught the wire will be

$i=\frac{V}{R}=\frac{V}{2\pi Na\rho }$

And we know that magnetic field produced by toroid inside the core of toroid is given by

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

Putting the value of $i$, we get

$B=\frac{{\mu }_{0}N}{2\pi r}\frac{V}{2\pi Na\rho }$

$\therefore V=\frac{4{\pi }^{2}Nra\rho B}{{\mu }_{0}N}$

Why this question ? Tip: It enhances the concept of Ohm's law and magnetic field in the core of toroid.