Question

A toroid has inner radius of $5\text{cm}$ and outer radius of $8\text{cm}$. A point $\text{P}$ is located inside the toroid at a distance of $5.5\text{cm}$ from the center of the toroid. The toroid carries a current $I$ and magnetic field at point $\text{P}$ is equal to $B$ due to this current. If the toroid suddenly shrinks such that its inner radius becomes $3.5\text{cm}$ and outer radius becomes $7\text{cm}$, keeping the current and number of turns same. What will be the value of magnetic field at point $\text{P}$ in the later case ?

• A. $B/2$
• B. $B$
• C. $B/3$
• D. $2B$

Answer 1

The correct option is B $B$

$\text{For case I:}$ At normal condition,

Given,
$r_1=5\text{cm},r_2=8\text{cm}r=5.5\text{cm}$

As, $r_1<r<r_2$, so the magnetic field at point $\text{P}$ is calculated by using,

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

$\text{For case II:}$ When toroid shrinks,

$r_1^{\prime }=3.5\text{cm},r_2^{\prime }=7\text{cm},r=5.5\text{cm}$

We can see that point $\text{P}$ still lies inside the toroid such that, $r_1<r<r_2$.

Therefore, the magnetic field will remains same at point $\text{P}$ as everything in the expression of magnetic field for point $\text{P}$ remains same.

Therefore, option $\left(B\right)$ is correct.