Question

Parallel conducting tracks, separated by 2.20 cm, run north and south. There is a uniform magnetic field of 1.20 T pointing upward (out of the page). A 0.0400-kg cylindrical metal rod is placed across the tracks and a battery is connected between the tracks, with its positive terminal connected to the east track. The current through the rod is 1.40 A. What is the magnitude of the magnetic force on the rod

Answer 1

Answer:

The magnitude of the magnetic force on the rod is 0.037 N.

Explanation:

The magnetic force is given by:

$F = qvBsin(\theta)$

Since the charge (q) is:

$q = I*t$

Where I is the current = 1.40 A, and t the time  

And the speed (v):

$v = \frac{L}{t}$

Where L is the tracks separation = 2.20 cm = 0.022 m

Hence, the magnetic force is:

$F = ILBsin(\theta)$

Where B is the magnetic field = 1.20 T and θ is the angle between the tracks and the magnetic field = 90°

$F = ILBsin(\theta) = 1.40*0.022*1.20*sin(90) = 0.037 N$

Therefore, the magnitude of the magnetic force on the rod is 0.037 N.

I hope it helps you!