A 30 -turn square coil with a mass of \(0.250 \mathrm{~kg}\) and a side length of \(0.200 \mathrm{~m}\) is hinged along a horizontal side and carries a 5.00 -A current. It is placed in a magnetic field pointing vertically downward and having a magnitude of \(0.00500 \mathrm{~T}\). Determine the angle that the plane of the coil makes with the vertical when the coil is in equilibrium. Use \(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\).

First, we need to identify the forces acting on the coil. There are two main forces to consider: the gravitational force and the magnetic force. The gravitational force acts vertically downward on the center of mass of the coil, and its magnitude is equal to the product of the mass of the coil and the acceleration due to gravity (mg). The magnetic force acts on each side of the coil, but only the forces acting on the top and bottom sides create a torque about the hinge. The magnitude of the magnetic force on each side can be calculated using the formula: \(F_B = BIL\), where B is the magnetic field, I is the current, and L is the length of the side.

The torque \(\tau\) is the turning force that causes the coil to rotate about the hinge. The torque due to the gravitational force and the magnetic force should be equal for the coil to be in equilibrium. The torque due to the gravitational force can be calculated as follows: \(\tau_g = Mg\frac{L}{2}\sin(\theta)\) where \(\theta\) is the angle the plane of the coil makes with the vertical, M is the mass of the coil, g is the acceleration due to gravity, and L/2 is the distance from the center of mass to the hinge. The torque due to the magnetic force can be calculated as: \(\tau_B = BINA\cos(\theta)\) where A is the area of the coil, and N is the number of turns.

For the coil to be in equilibrium, the torques due to the gravitational force and the magnetic force must be equal: \(\tau_g = \tau_B\) Substitute the expressions for \(\tau_g\) and \(\tau_B\) and solve for the angle \(\theta\): \(Mg\frac{L}{2}\sin(\theta) = BINAL\cos(\theta)\) Divide both sides by \(\frac{ML}{2}\): \(g\sin(\theta)=2BINA\) Now, divide both sides by \(g\): \(\sin(\theta)=\frac{2BINA}{g}\) Finally, take the inverse sine of both sides to find the angle \(\theta\): \(\theta = \arcsin\left(\frac{2BINA}{g}\right)\) Substitute the given values for B=0.00500 T, I=5.00 A, N=30, A=\((0.200\,\mathrm{m})^2\), and \(g=9.81\,\mathrm{m/s^2}\): \(\theta = \arcsin\left(\frac{2(0.00500\,\mathrm{T})(5.00\,\mathrm{A})(30)(0.200\,\mathrm{m})^2}{9.81\,\mathrm{m/s^2}}\right)\) Now, calculate the value of \(\theta\): \(\theta \approx 18.5^{\circ}\) Hence, the angle that the plane of the coil makes with the vertical when the coil is in equilibrium is approximately \(18.5^{\circ}\).