Question

a solenoid is 30.0 cm long, has a radius of 1.00 cm, has 650 turns and is carrying 0.750 a of current. there is a magnetic field with a magnitude of 0.900 t that is pointing vertically as shown in the figure. if the solenoid feels a magnitude of torque equal to 0.0380 nm, what is the angle between the axis of the solenoid (the dashed line) and horizontal?

Answer 1

The angle between the axis of the solenoid and horizontal is 25.2 degrees.

Step-by-step explanation:

To find the angle between the axis of the solenoid and horizontal, we can use the formula:

Torque = Magnetic Field * Current * Area * Sin(theta)

We know the torque (0.0380 Nm), magnetic field (0.900 T), current (0.750 A), and area (π * r2). We can rearrange the formula to solve for theta:

Theta = arcsin(Torque / (Magnetic Field * Current * Area))

Plugging in the numbers, we get:

Theta = arcsin(0.0380 Nm / (0.900 T * 0.750 A * π * (0.01 m)2))

Calculating this gives us:

Theta = 25.2 degrees