Question

A coil wrapped around a toroid has inner radius of $10.0\text{cm}$ and an outer radius $20.0\text{cm}$. If the wire wrapped makes $600\text{turns}$ and carries current of $10.0\text{A}$. Choose the correct alternatives.
$[B_{max}$ and $B_{min}$ are maximum and minimum magnetic fields inside the toroid$]$

• A. $B_{max}=12\text{mT}$
• B. $B_{max}=6\text{mT}$
• C. $B_{min}=6\text{mT}$
• D. $B_{min}=3\text{mT}$

Answer 1

Answer: (A), (C)

$B_{min}=6\text{mT}$

Given:
Current, $I=10\text{A}$
inner radius of toroid, $r_1=10\times {10}^{-2}\text{m}$
Outer radius of toroid, $r_2=20\times {10}^{-2}\text{m}$
Total number of turns, $N=600$

As we know that magnetic field inside the toroid at distance $r$ is

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

So, at $r=r_1$, $B$ will be maximum

$\Rightarrow B_{max}=\frac{{\mu }_{0}NI}{2\pi r_1}=\frac{(4\pi \times {10}^{-7})\times 600\times 10}{2\pi \times (10\times {10}^{-2})}$

$\therefore B_{max}=12\times {10}^{-3}\text{T}=12\text{mT}$

Similarly, minimum magnetic field will be at $r=r_2$.

$\Rightarrow B_{min}=\frac{{\mu }_{0}NI}{2\pi r_2}=\frac{(4\pi \times {10}^{-7})\times 600\times 10}{2\pi \times (20\times {10}^{-2})}$

$\therefore B_{min}=6\times {10}^{-3}\text{T}=6\text{mT}$

Therefore, options $\left(A\right)$ and $\left(C\right)$ are the correct answer.