A 0.320 kg,25-turn square coil has a side of 0.210 m,is hinged along a horizontal side and carries 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 current in the coil.Use g = 9.81 m/s².

Question

Quizplus · Accepted Answer

The torque due to the magnetic field on the coil is balanced by the gravitational torque at equilibrium. The magnetic torque ($\tau_m$) is given by $\tau_m = NIAB\sin(\theta)$, 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 ($\tau_g$) is given by $\tau_g = mg\frac{l}{2}\cos(\theta)$, where $m$ is the mass of the coil, $g$ is the acceleration due to gravity, $l$ is the length of a side of the square coil, and $\theta$ is the same angle. Setting $\tau_m = \tau_g$ and solving for $I$ gives the correct current.

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 $\circ$