Question

A galvanometer coil has $500$ turns, and each turn has an average area of $3\times {10}^{-4}{\text{m}}^2$. If a torque of $1.5\text{Nm}$ is required to keep this coil parallel to a magnetic field when a current of $0.5\text{A}$ is flowing through it, the strength of the field \$\text{(in T)}$ is .

Answer 1

Given,
$\text{Area of the coil,}A=3\times {10}^{-4}{\text{m}}^2$
$\text{Number of turns in the coil,}N=500$
$\text{Current in the coil,}I=0.5\text{A}$

$\text{Torque,}\tau =|\overrightarrow{M}\times \overrightarrow{B}|$

$\Rightarrow \tau =NiAB\sin \left({90}^{\circ }\right)=NiAB$

$\Rightarrow B=\frac{\tau }{NiA}$

$\Rightarrow B=\frac{1.5}{500\times 0.5\times 3\times {10}^{-4}}=20\text{T}$

Hence, $20$ is the correct answer.