Minimum magnetic field that will suspend

Question

A wire of length 19.7 cm and mass 6.46 grams carries a current of 0.98 A. What is the minimum magnetic field that will suspend it?

Answer 1

For this problem you need to know the force on a current carrying wire placed in a magnetic field: F = (I)(L) x B, where I is the current, L is the length, and B is the magnetic field. Note the magnitude of the cross product is F = ILBsinθ.

To be suspended, the sum of the forces on the wire must be 0. We take it that gravity is acting downwards and the force due to the magnetic field is upwards:

∑F = F_{B -} F_{G =} 0

This tells you what the magnitude of F_B needs to be.

Because the direction of the current is not specified, let's assume that we can orient the wire whichever way and the same for the magnetic field. We would do it such that the cross product points upwards, but since this question doesn't care about direction, we just need to find the magnitude.

To minimize the other terms in the magnitude (in this case B because I and L are given), we should maximize the cross product, which is 1 because thats the greatest value of sinθ (note the max value occurs when the direction of the current and the magnetic field are perpendicular). Knowing these bits should allow you to plug in the numbers and solve the problem.