A solenoid is wound with 250 turns per centimeter. An outer layer of insulated w

Question

A solenoid is wound with 250 turns per centimeter. An outer layer of insulated wire with 200 turns per centimeter is wound over the solenoid's first layer of wire.

When the solenoid is operating, the inner coil carries a current of 10 A and the outer coil carries a current of 15 A in the direction opposite to that of the current in the inner coil.(Figure 1)

What is the direction of the magnetic field at the center for this configuration?

Right, left, downward, or upward?

What is the magnitude of the magnetic field at the center of the doubly wound solenoid?

Explanation / Answer

The magnetic field in a solenoid is

B = mu0 * I * n

[to the right]

--------------------------------------------------

magnetic field from the inner layer

Binner = 4pi * 10-7 * 10 * [250 /10-2]

= 0.314

Bouter = 4pi * 10-7 * 15 * [200 /10-2]

= 0.377

overall magnetic field = 0.377 - 0.314

= 0.0629 T

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