A galvanometer coil has 500 turns and each turn has an average area

Question:

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

Solution:

(20)

Given,

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

Number of turns in the coil, $N=500$

Current in the coil, $I=0.5 \mathrm{~A}$

Torque $\tau=|\vec{M} \times \vec{B}|=N i A B \sin \left(90^{\circ}\right)=N i A B$

$\Rightarrow B=\frac{\tau}{N i A}=\frac{1.5}{500 \times 0.5 \times 3 \times 10^{-4}}=20 T$

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