A 25-turn square coil has a side of 0.210 m,is hinged along a horizontal side and carries a 6.00-A current.When placed in a vertical magnetic field pointing down and having a magnitude of 4.00 mT,the plane of the coil makes a 2.3 $\circ$ angle with the vertical when the coil is in equilibrium.Determine the mass of the coil.Use g = 9.81 m/s².

Question

Quizplus · Accepted Answer

The torque on the coil is given by:

τ=NIABsinθ

where N is the number of turns, I is the current, A is the area of the coil, B is the magnetic field, and θ is the angle between the plane of the coil and the magnetic field.

The magnitude of the torque is balanced by the gravitational torque:

τ=mg(L/2)sinφ

where m is the mass of the coil, L is the side length of the coil, and φ is the angle between the plane of the coil and the horizontal.

Setting these two torques equal and solving for the mass, we get:

m=(2NIB/gL)sinθ/sinφ

Plugging in the given values, we get:

m=(2*25*6.00*0.210*0.00400/9.81*0.210)sin2.3°/sin90°≈0.640 kg

Therefore, the answer is choice (B).

A 25-Turn Square Coil Has a Side of 0 ∘\circ∘ Angle with the Vertical When the Coil Is in Hinged

A 25-Turn Square Coil Has a Side of 0 $\circ$ Angle with the Vertical When the Coil Is in Hinged