Question

A conductor wire with length of 0.20 m and weight of 1.47 N is oriented perpendicularly to a magnetic field. If the current of 1.75A flows to the north, the net force on the wire is measured using a balance and found to be zero. What is the magnitude and direction of the magnetic field?

Answer 1

1. Magnitude: 4.4 T

The net force on the wire is zero, therefore the magnetic force must be equal to the weight of the wire:

$F_M = W = 1.47 N$

Where the magnetic force is

$F_M = IBL sin \theta$

where

I = 1.75 A is the current

B is the magnetic field

L = 0.20 m is the length of the wire

$\theta=90^{\circ}$ is the angle between the direction of L and B

Solving the formula for B, we find the magnitude of the magnetic field:

$B=\frac{F_M}{ILsin \theta}=\frac{1.47 N}{(1.75 A)(0.20 m)(sin 90^{\circ})}=4.2 T$

2. Direction: westward

The direction is given by the right's hand rule. We know that:

- thumb --> magnetic force (upward)

- index finger --> current (north)

- middle finger --> direction of magnetic field: so, it must be westward.

- middle