A uniform straight conducting rod of mass 200 g and cross-sectional area 2 mm ^2 is pivoted freely at its upper end and hangs vertically. Its lower end dips into a small pool of liquid metal to complete an electrical circuit. A battery of voltage V is connected across the rod. A uniform horizontal magnetic field of 0.1 T is applied perpendicular to the plane of the rod's swing. When the current flows, the rod deflect

Step-by-step solution

Let the length of the rod be L . The magnetic field B is perpendicular to the plane of the swing, so it is always perpendicular to the length of the rod. The magnetic force on a small element dy at a distance y from the pivot is dF = IBdy . This force acts perpendicular to the rod. The torque due to this magnetic force about the pivot is: _B = ₀^ L (IBdy)y = IBL^2 2 The torque due to gravity acts at the center of mass ( L/2 from the pivot) and is given by: _g = mg ( L 2 ) In equilibrium, these torques balance each