# Current in wire loop inside a solenoid

• gills

## Homework Statement

A 2.1m -long solenoid is 13.0cm in diameter and consists of 2200 turns of wire. The current in the solenoid is increasing at the rate of 1.0 kA/s .

## The Attempt at a Solution

Any help to point me in the right direction would be great.

I'm thinking that i need to find the magnetic field inside that of a larger solenoid. Then i somehow manage to put that into faradays law equation.

bump^^

That is exactly what you have to do. Good thinking.

HINT: How can you relate the current in the solenoid to the B-Field produced?

That is exactly what you have to do. Good thinking.

HINT: How can you relate the current in the solenoid to the B-Field produced?

how do i treat the current since it is given per unit time? That is throwing me off.

B(solenoid) = $$\mu$$$$_{0}$$*n*I ---> ??

ok this is what I've got

dB/dt = $$\mu$$$$_{0}$$*n*$$\frac{dI}{dt}$$

ok this is what I've got

dB/dt = $$\mu$$$$_{0}$$*n*$$\frac{dI}{dt}$$

ok, got the first part.

What about when there's a loop that is larger than the solenoid? Would the current in that loop be zero since the magnetic field outside of solenoid is almost neglible?

The diameter of the larger loop is 19.5cm

ok, got the first part.

What about when there's a loop that is larger than the solenoid? Would the current in that loop be zero since the magnetic field outside of solenoid is almost neglible?

The diameter of the larger loop is 19.5cm

ok, nevermind on that one. Would i just be changing the area of the loop from the equation i derived in the first problem or is there more to it?

ok, nevermind on that one. Would i just be changing the area of the loop from the equation i derived in the first problem or is there more to it?

Yes, the only difference when the loop is bigger than the solenoid is the area of the loop. (The magnetic flux still goes through the loop if it's bigger than the solenoid.)

Keep working at it, you seem to be making some nice progress on this problem. Keep it up! If you have any more questions I'll answer them as soon as I can.