Question

A circular coil of $20$ turns having radius $10\text{cm}$ is placed in a uniform magnetic field of $0.2\text{T}$ normal to the plane of the coil, If a current of $5\text{A}$ passes through the coil, then the torque experienced by the coil in $\left(\text{Nm}\right)$

• A. $0.314$
• B. $0.928$
• C. $0.628$
• D. $0.528$

Answer 1

The correct option is C $0.628$

Given,

$N=20\text{turns};r=10\text{cm}=10\times {10}^{-2}\text{m}B=0.2\text{T};A=\pi r^2=\pi (10\times {10}^{-2}{)}^2=\pi \times {10}^{-2}{\text{m}}^{2}i=5\text{A};\theta ={90}^{\circ }$

Torque experienced by the coil is,

$\overrightarrow{\tau }=\overrightarrow{M}\times \overrightarrow{B}=MB\sin \theta$

$\Rightarrow \tau =NiAB\sin \theta$

$=20\times 5\times \pi \times {10}^{-2}\times 0.2\times 1$

$=2\times 3.14\times {10}^{-1}$

$=0.628\text{Nm}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.