Question

Suppose a straight wire with a length of 2.0 m runs perpendicular to a magnetic field with a magnitude of 38 T. What current would have to pass through the wire in order for the magnetic force to equal the weight of a student with a mass of 75 kg?

Answer 1

Answer:

9.67 A

Explanation:

We want the magnetic force on the wire to be equal to the weight of a student of mass m=75 kg. The magnetic force on the wire is given by

$F=ILB sin \theta$

where

I is the current in the wire

L = 2.0 m is the length of the wire

B = 38 T is the magnetic field

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

While the weight of the student is

$W=mg=(75 kg)(9.8 m/s^2)=735 N$

where g=9.8 m/s^2 is the acceleration due to gravity.

The problem can be solved by equalizing the two forces: W=F. So we can write

$ILB sin \theta=W$

And solving for I, the current, we find

$I=\frac{W}{BLsin\theta}=\frac{735 N}{(38 T)(2.0 m)(sin 90^{\circ})}=9.67 A$