A 0.320 kg,25-turn square coil has a side of 0.210 m,is hinged along a horizontal side and carries a 6.0-A current.When placed in a vertical magnetic field pointing down,the plane of the coil makes a 7.3 $\circ$ angle with the vertical when the coil is in equilibrium.Determine the magnitude of the magnetic field.Use g = 9.81 m/s².

Question

Quizplus · Accepted Answer

The torque ($\tau$) due to the magnetic field on the coil is balanced by the gravitational torque at equilibrium. The magnetic torque is given by $\tau = NIAB\sin(\theta)$, where $N$ is the number of turns, $I$ is the current, $A$ is the area of the coil, $B$ is the magnetic field strength, and $\theta$ is the angle with the vertical. The gravitational torque is $\tau_g = mgL/2\cos(\theta)$, where $m$ is the mass of the coil, $g$ is the acceleration due to gravity, and $L/2$ is the distance from the hinge to the coil's center of mass (half the side length of the square coil). Setting these equal and solving for $B$ gives the correct magnitude of the magnetic field.

A 0320 Kg,25-Turn Square Coil Has a Side of 0 ∘\circ∘

A 0320 Kg,25-Turn Square Coil Has a Side of 0 $\circ$