A loop with a resistance of 4.0 is pushed to the left at a constant speed of 2.0 m/s by a 24 N force.At the instant shown in the figure,the loop is partially in and partially out of a uniform magnetic field.An induced current flows from left to right through the resistor.The length and width of the loop are 2.0 m and 1.0 m,respectively.

-Determine the magnitude of the uniform magnetic field.

Question

A loop with a resistance of 4.0  is pushed to the left at a constant speed of 2.0 m/s by a 24 N force.At the instant shown in the figure,the loop is partially in and partially out of a uniform magnetic field.An induced current flows from left to right through the resistor.The length and width of the loop are 2.0 m and 1.0 m,respectively.
 
-Determine the magnitude of the uniform magnetic field.

Quizplus · Accepted Answer

The equation for the induced current in the loop is given by:

I = (F*B*w*l)/(R*v)

where F is the applied force, B is the magnetic field, w and l are the width and length of the loop, R is the resistance of the loop, and v is the velocity of the loop.

Substituting the given values, we get:

I = (24*B*1*2)/(4*2) = 6B

The direction of the induced current (from left to right) shows that the magnetic field and the applied force are in opposite directions (by Lenz's law). Therefore, we can use the right-hand rule to determine that the magnetic field is into the page.

Using the magnitude of the induced current (6B) and the speed of the loop (2 m/s), we can solve for the magnitude of the magnetic field:

6B = (24*2)/4
B = 2 T

Therefore, the answer is choice B (3.5 T).

A Loop with a Resistance of 4