Question

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 \(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\)

Short answer

Answer: The angle is approximately \(29.4^{\circ}\).

Step-by-step solution

Write the equation for magnetic torque

The magnetic torque, \(\tau_{m}\), acting on the current-carrying coil can be calculated using the following formula: \(\tau_{m} = nBAI\sin{\theta}\) where n is the number of turns, B is the magnetic field, A is the area of the coil, I is the current flowing through the coil, and \(\theta\) is the angle between the normal to the plane of the coil and the magnetic field direction.

Calculate the area of the coil

The coil is square-shaped, so the area can be calculated using the formula: \(A = l^{2}\) where \(l\) is the side length of the coil. Substitute the value of \(l\) to find the area: \(A = (0.200 \mathrm{~m})^2 = 0.0400 \mathrm{~m}^2\)

Write the equation for gravitational torque

The gravitational torque, \(\tau_{g}\), acting on the coil can be calculated using the formula: \(\tau_{g} = mgr\sin{\alpha}\) where m is the mass of the coil, g is the acceleration due to gravity, r is the distance from the hinge to the center of mass, and \(\alpha\) is the angle between the line connecting hinge and center of mass and the vertical. Since the coil is symmetrical and hinged along one of its horizontal sides, the distance r is equal to half the side length, \(r = \frac{l}{2}\), and the angle \(\alpha\) is equal to the angle \(\theta\) (the angle we want to find).

Balance the torques and solve for the angle

In equilibrium, the net torque on the coil is zero, so the magnetic torque and gravitational torque must balance each other: \(\tau_{m} = \tau_{g}\) Substitute the formulas for \(\tau_{m}\) and \(\tau_{g}\): \(nBAI\sin{\theta} = mgr\sin{\alpha}\) As mentioned earlier, \(\alpha = \theta\). So, the equation becomes: \(nBAI\sin{\theta} = mgr\sin{\theta}\) Now, we can divide both sides by \(\sin{\theta}\) and plug in the known values: \(30 \cdot 0.00500 \mathrm{~T} \cdot 0.0400 \mathrm{~m}^2 \cdot 5.00\mathrm{~A} = 0.250\mathrm{~kg} \cdot [\frac{1}{2}(0.200 \mathrm{~m})] \cdot 9.81\frac{\mathrm{m}}{\mathrm{s}^2}\) Solve for \(\sin{\theta}\): \(\sin{\theta} = \frac{0.250\mathrm{~kg} \cdot [\frac{1}{2}(0.200 \mathrm{~m})] \cdot 9.81\frac{\mathrm{m}}{\mathrm{s}^2}}{30 \cdot 0.00500 \mathrm{~T} \cdot 0.0400 \mathrm{~m}^2 \cdot 5.00\mathrm{~A}} \approx 0.489\)

Find the angle

Now, to find the angle \(\theta\), take the inverse sine of 0.489: \(\theta = \sin^{-1}(0.489) \approx 29.4^{\circ}\) Thus, the plane of the coil makes an angle of approximately \(\boldsymbol{29.4^{\circ}}\) with the vertical when the coil is in equilibrium.