A solenoid is wound with 290 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.

What is the direction of the magnetic field at the center for this configuration?
a)upward
b)downward
c)to the left
d)to the right

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

Without a figure, we have no way to relate up, down, left and right to the configuration and orientation of the solenoid.

There should be a formula for the magnetic field inside a solenoid in your textbook.

Downward

To determine the direction of the magnetic field at the center of the doubly wound solenoid, you can use the right-hand rule. The right-hand rule states that if you point your right thumb in the direction of the current flow and wrap your fingers around the wire, your fingers will curl in the direction of the magnetic field.

For the inner coil, the current flows in the clockwise direction (given that the outer coil carries a current in the opposite direction). So, if you curl your fingers around the inner coil in the clockwise direction, your thumb will point upward. This means that the magnetic field due to the current in the inner coil is directed upward.

For the outer coil, the current flows in the counterclockwise direction (opposite to the current in the inner coil). So, if you curl your fingers around the outer coil in the counterclockwise direction, your thumb will point downward. This means that the magnetic field due to the current in the outer coil is directed downward.

Now, determine the net magnetic field at the center of the doubly wound solenoid by adding the magnetic fields due to the inner and outer coils together. Since the magnetic fields are in opposite directions, the net magnetic field will be the difference between the two magnitudes.

Given that the inner coil carries a current of 10 A and the outer coil carries a current of 15 A, you can use Ampere's law to calculate the magnetic field magnitude at the center:

B = (μ₀ * N * I) / d

Where B is the magnetic field magnitude, μ₀ is the permeability of free space (4π x 10^-7 T⋅m/A), N is the number of turns per unit length (turns per centimeter), I is the current, and d is the distance from the center of the solenoid.

For the inner coil, N = 290 turns/cm and I = 10 A. For the outer coil, N = 200 turns/cm and I = 15 A.

By substituting the values into the formula, you can calculate the respective magnetic field magnitudes:

B_inner = (4π x 10^-7 T⋅m/A) * (290 turns/cm) * (10 A) / d
B_outer = (4π x 10^-7 T⋅m/A) * (200 turns/cm) * (15 A) / d

Finally, subtract the magnitude of the outer coil from the magnitude of the inner coil to find the net magnetic field at the center of the doubly wound solenoid:

B_net = B_inner - B_outer