Question

A circular loop of radius $a=7\text{cm}$, carrying a current $i=\frac{2}{7}\text{A}$, is placed in a two-dimensional magnetic field. The centre of the loop coincides with the centre of the field. The strength of the magnetic field at the periphery of the loop is $B$. Find $B$ if the magnetic force on the wire is $2\pi \text{N}$.

• A. $20\text{T}$
• B. $25\text{T}$
• C. $7\text{T}$
• D. $50\text{T}$

Answer 1

The correct option is D $50\text{T}$

Given:

Radius of the loop, $a=7\text{cm}$,

So, length of the loop,

$l=2\pi a=2\pi \times 7=14\pi \times {10}^{-2}\text{m}$,

Current in the loop, $i=\frac{2}{7}\text{A}$

Magnetic force on the small segment of the loop,

$\overrightarrow{dF}=i(\overrightarrow{d}l\times \overrightarrow{B})\Rightarrow dF=idlB\sin \theta$


As angle between $\overrightarrow{dl}$ and magnetic field $\overrightarrow{B}$ is ${90}^{\circ }$, so,

$dF=idlB\sin {90}^{\circ }=idlB$

For whole loop,

$F=ilB$

Substituting the values,

$\Rightarrow 2\pi =\frac{2}{7}\times 14\pi \times {10}^{-2}\times B$

$\therefore B=50\text{T}$

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