A simple galvanometer is made from a coil that consists of \(N\) loops of wire of area \(A .\) The coil is attached to a mass, \(M\), by a light rigid rod of length \(L\). With no current in the coil, the mass hangs straight down, and the coil lies in a horizontal plane. The coil is in a uniform magnetic field of magnitude \(B\) that is oriented horizontally. Calculate the angle from the vertical of the rigid rod as a function of the current, \(i\), in the coil.

To find the angle from the vertical, we need to analyze the torques acting on the system. The two forces acting on the system are the gravitational force acting on the mass, and the magnetic force acting on the coil. The gravitational force creates a torque about the axis of rotation that opposes the angle, and the magnetic force creates a torque that helps the angle to increase.

The gravitational force acting on the mass creates a torque about the axis of rotation. This torque can be expressed as: \(τ_g = MgLsin(θ)\) where \(τ_g\) is the torque due to gravity, \(M\) is the mass, \(g\) is the acceleration due to gravity, \(L\) is the length of the rigid rod, and \(θ\) is the angle from the vertical.

The magnetic force acting on the coil creates a torque about the axis of rotation. To calculate this torque, we need to first find the magnetic force acting on each segment of the coil. The magnetic force on a current loop can be expressed as: \(F_m = NiABsin(α)\) where \(F_m\) is the magnetic force, \(N\) is the number of loops in the coil, \(i\) is the current in the coil, \(A\) is the area of the coil, \(B\) is the magnitude of the magnetic field, and \(α\) is the angle between the magnetic field and the plane of the coil. Since the magnetic field is horizontal and the coil lies in a horizontal plane, \(α = 90°\). Therefore, the magnetic force acting on each segment of the coil is: \(F_m = NiAB\) The torque due to this magnetic force can be expressed as: \(τ_m = F_mLcos(θ)\) where \(τ_m\) is the torque due to magnetic force.

At equilibrium, the total torque acting on the system is zero. Therefore, we can set up the following equation: \(τ_g = τ_m\) Substituting the values of torques due to gravitational and magnetic forces, we get: \(MgLsin(θ) = NiABLcos(θ)\) Now we need to solve for \(θ\): \(sin(θ)cos(θ)=\frac{NiAB}{Mg}\) Using the trigonometric identity \(2sin(θ)cos(θ)=sin(2θ)\), we can rewrite the equation as: \(sin(2θ) = \frac{2NiAB}{Mg}\) Now solve for \(θ\): \(θ = \frac{1}{2} arcsin(\frac{2NiAB}{Mg})\) This expression gives the angle of the rigid rod from the vertical as a function of the current in the coil.