Question

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

Answer 1

Step 1. Given data:

Area of galvanometer coil, $A=3\times {10}^{-4}{m}^2$

Current flowing in the coil, $I=0.5A$

Number of turns, $N=500$

Torque required, $𝜏=1.5Nm$

Magnetic field strength, $B=?$

Step 2. Calculating magnetic field strength of the galvanometer coil

$𝜏=B\times I\times N\times A$

Putting values in the above equation, we get

$\begin{array}{rcl}& \Rightarrow & 1.5=B\times 0.5\times 500\times 3\times {10}^{-4}\\ & \Rightarrow & B=\frac{{10}^4}{500}\\ \therefore B& =& 20T\end{array}$

Hence, Magnetic field strength of the coil is, $\begin{array}{rcl}\mathbf{}\mathit{B}\mathbf{}& \mathbf{=}& \mathbf{}\mathbf{20}\mathbf{}\mathit{T}\end{array}$