Question

Two identical wires are used to make toroids $\text{A}$ and $\text{B}$ of cross-sectional radius $a$ and $b$ respectively. Average radii of toroid $\text{A}$ and $\text{B}$ are $r$ and $2r$ respectively. Find the ratio of magnetic field of $\text{A}$ and $\text{B}$, if they are connected across a battery of emf $E$ and $3E$ respectively.
[ wire is closely wounded]

• A. $\frac{3b}{2a}$
• B. $\frac{3a}{2b}$
• C. $\frac{2a}{3b}$
• D. $\frac{2b}{3a}$

Answer 1

The correct option is D $\frac{2b}{3a}$

Let the length of the wire be $L$ and resistance be $R$.

Current in $\text{A}$, $i_A=\frac{V_A}{R}=\frac{E}{R}$

Similarly, current in $\text{B}$, $i_B=\frac{3E}{R}$

Length of one turn of $\text{A}$, $l_A=2\pi a$

Similarly, $l_B=2\pi b$

So, total number of turns of $\text{A}$,
$N_A=\frac{L}{l_A}=\frac{L}{2\pi a}$

Similarly, $N_B=\frac{L}{l_B}=\frac{L}{2\pi b}$

Now using formula of magnetic field inside the core of toroid,

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

So, for toroid $\text{A}$,
$B_A=\frac{{\mu }_0{N}_A{i}_A}{2\pi r_A}=\frac{{\mu }_0\left(\frac{L}{2\pi a}\right)\times \frac{E}{R}}{2\pi \times r}$

Similarly,
$B_B=\frac{{\mu }_0{N}_B{i}_B}{2\pi r_B}=\frac{{\mu }_0\left(\frac{L}{2\pi b}\right)\times \frac{3E}{R}}{2\pi \times \left(2r\right)}$

$\therefore \frac{B_A}{B_B}=\frac{2b}{3a}$

Hence, option $\left(d\right)$ is right.