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A coil of 1000 turns of wire has a radius of 12 cm and carries a counterclockwise current of 15A. If it is lying flat on the ground, and the Earth's magnetic field points due north, has a magnitude of 5.8 x 10-5 T, and makes a downward angle of 25° with the vertical, what is the torque on the loop?

Answer :

Answer:

torque is 1.7 * [tex]10^{-2}[/tex] Nm

Explanation:

Given data

turns n = 1000 turns

radius r  = 12 cm

current I = 15A

magnitude B = 5.8 x 10^-5 T

angle θ = 25°

to find out

the torque on the loop

solution

we know that torque on the loop is

torque = N* I* A*B* sinθ

here area A = πr² = π(0.12)²

put all value

torque = N* I* A*B* sinθ

torque = 1000* 15* π(0.12)² *5.8 x 10-5 * sin25

torque = 0.0166 N m

torque is 1.7 * [tex]10^{-2}[/tex] Nm

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