Question

A 50-turn circular coil with radius 4 cm is placed in magnetic field of 6000 G. An amount of 800 mA current passes through the circular coil and the area vector of plane makes 40o with the magnetic field. Torque (τ) acting on the coil is  ​

Answer 1

Explanation:

n=50,r=0.02mn=50,r=0.02m,

I=5AandB=0.20TI=5AandB=0.20T

τismaxiμmwhensinθ=90∘τismaxiμmwhensinθ=90∘

τmax=niabsin90∘=mbτmax=niabsin90∘=mb

=50×5×3.14×4×10−4×2×10−1=50×5×3.14×4×10-4×2×10-1

=6.28×10−2Nm=6.28×10-2Nm

Given τ=12×τmaxτ=12×τmax

⇒sinsinθ=12⇒sinsinθ=12 or sinθ=30∘sinθ=30∘

=∠betweenareavar→randmag≠ticfield=∠betweenareavar→randmag≠ticfield.

So angle between magnetic field and the plane of the coil

=90∘−30∘=60∘=90∘-30∘=60∘.

HOPE IT HELPS

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