Find (a) the direction and (b) magnitude of the electric current to levitate the copper rod’s gravitation force.

Question

A copper rod 0.570 m long and with a mass 0.05900kg is suspended from two thin wire. At right angle to the rod is a uniform magnetic field of 0.670 T pointing into the page. Find (a) the direction and (b) magnitude of the electric current to levitate the copper rod’s gravitation force.

Answer 1

This is a case of a linear conductor carrying current in a magnetic field. The most fundamental equation is the force on a charge moving through a magnetic field is given by

F = q v X B

This equation leads to the commonly used formula

F = iLB Sin θ

where i is the current through the conductor

L is the length of the conductor

B is magnetic field strength

θ is the angle between i and B directions

The direction of the force is found by the right-hand rule. We want this force to act vertically up in order to counter the force of gravity. So, if the current flow is from left to right and the magnetic field is into the page, then the force will be facing vertically upwards. That force must balance the weight of the rod.

Since θ = 90º

Mg = iLB

You can calculate the current from this equation.