A coil of radius R = 5 cm lies inside a long solenoid which carries a current I₁ = 10 A and has 800 turns/m.The coil carries a current I₂ = 4 A and the normal of the plane of the coil is at an angle $\theta$ = 30 $\circ$ with the axis of the solenoid.The magnitude of the torque on the coil is A)1.56 $\times$ 10^-4 N . m B)2.74 $\times$ 10^-4 N . m C)5.03 $\times$ 10^-⁵ N . m D)3.16 $\times$ 10^-4 N . m E)none of the above

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Quizplus · Accepted Answer

The torque on the coil is given by:$$	au = NIAB\sin	heta$$where:* N is the number of turns in the coil* I is the current in the coil* A is the area of the coil* B is the magnetic field strength* θ is the angle between the normal to the plane of the coil and the magnetic fieldIn this case, we have:* N = 1* I = 4 A* A = πR^2 = π(0.05 m)^2 = 7.85 x 10^-3 m^2* B = μ0nI1 = (4π x 10^-7 T·m/A)(800 turns/m)(10 A) = 0.0251 T* θ = 30°Plugging these values into the equation, we get:$$	au = (1)(4 A)(7.85 x 10^-3 m^2)(0.0251 T)\sin(30°) = 1.56 x 10^-4 N \cdot m$$Therefore, the correct answer is A.

A Coil of Radius R = 5 Cm Lies Inside θ\thetaθ

A Coil of Radius R = 5 Cm Lies Inside $\theta$