Question

A thin aluminum rod lies along the x-axis and has current of I = 16.0 A running through it in the +x-direction. The rod is in the presence of a uniform magnetic field, perpendicular to the current. There is a magnetic force per unit length on the rod of 0.113 N/m in the −y-direction.
(a) What is the magnitude of the magnetic field (in mT) in the region through which the current passes?
(b) What is the direction of the magnetic field in the region through which the current passes?

Answer 1

Answer:

a) The magnitude of the magnetic field = 7.1 mT

b) The direction of the magnetic field is the +z direction.

Explanation:

The force, F on a current carrying wire of current I, and length, L, that passes through a magnetic field B at an angle θ to the flow of current is given by

F = (B)(I)(L) sin θ

F/L = (B)(I) sin θ

For this question,

(F/L) = 0.113 N/m

B = ?

I = 16.0 A

θ = 90°

0.113 = B × 16 × sin 90°

B = 0.113/16 = 0.0071 T = 7.1 mT

b) The direction of the magnetic field will be found using the right hand rule.

The right hand rule uses the first three fingers on the right hand (the thumb, the pointing finger and the middle finger) and it predicts correctly that for current carrying wires, the thumb is in the direction the wire is pushed (direction of the force; -y direction), the pointing finger is in the direction the current is flowing (+x direction), and the middle finger is in the direction of the magnetic field (hence, +z direction).