Question

For the magnetic field at some random angle to the plane of the small coil, draw a picture showing only the small coil, a vector giving the direction of the magnetic field, the area vector for the coil, and the angle between the magnetic field and the area vector. Write an equation for the magnetic flux through the small coil at an instant of time when the area vector is at some angle to the magnetic field. Write an expression that shows how the magnetic flux through the small coil changes as it turns with a constant angular speed.

Answer 1

Answer:

Solution attached below

Explanation:

Answer 2

A) An equation for the magnetic flux through the small coil at an instant of time when the area vector is at some angle to the magnetic field is;

Φ = BA cos θ

B) An expression that shows how the magnetic flux through the small coil changes as it turns with a constant angular speed is;

dΦ/dt = -BAω sinθ

I have attached an image showing a vector giving the direction of the magnetic field, the area vector and the angle between the magnetic field and the area vector.

A) The general formula for Magnetic flux is;

Φ = BA cos θ

Where;

B is magnetic field

A is area

θ is angle between magnetic field and the normal to the area

B) We are told that the magnetic flux changes as it turns with a constant angular speed.

Thus, we will differentiate both sides with respect to t.

dΦ/dt = -BA sinθ (dθ/dt)

Where (dθ/dt) is angular speed written as ω

Thus, we now have;

dΦ/dt = -BAω sinθ

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