A circular, 20-turn coil of radius 5.0 cm is oriented in such a way that its axis makes a 30

Question

Question

Answer 1

Answer #1

torque on a body is defined by the equation, T = NiAB sin theta

where N # of turns

i is current

A is area = here = pi r^2 = 3.14* 0.05*0.05 = 0.00785 m^2

B is magnetic field

theta = 30 deg, sin 30 = 0.5

so T = 20 * 2.5* 0.00785 * 0.15*0.5

T = 0.0294 Nm

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Answer 2

Answer #2

Area

A=pi*r²=pi*0.05²=7.854*10^-3 m²

Torque

T=NIABSin(o)

T=20*2.5*(7.854*10^-3)*0.15*sin30

T=0.02945 N-m

Answer 3

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