How Does Rotating a Semicircular Wire Loop Affect Magnetic Flux?

[rcmango]

Homework Statement

A loop of a wire has the shape shown in the drawing. The top part of the wire is bent into a semicircle of radius r = 0.24 m.

The normal to the plane of the loop is parallel to a constant magnetic field ( = 0°) of magnitude 0.75 T. What is the change in the magnetic flux that passes through the loop when, starting with the position shown in the drawing, the semicircle is rotated through half a revolution?
Wb in weber



here is a pic: http://img213.imageshack.us/img213/4917/75104403bx0.png

Homework Equations

The Attempt at a Solution

not sure where the time comes from, how to figure that out from the diagram.

also, are A0 = x0*L

that is the area swept out in time t.

need help finding the variables for this one.


also, i know that I need to find the angle A and B make at time t.

I need help on how to do this.

 

[Mindscrape]

So what you probably want to do in your imagination is rotate the loop and figure out how the area vector's normal changes. After all, flux will only go through the parts of the loop that are parallel to the magnetic field. How do the components of the area normal vary in time if we call the angular velocity of rotation omega?

The math formalism you want is essentially

$$\Phi = AB\matbf{\hat{n}}(t)$$

 

[Moetasim]

Dear student,

The change in magnetic flux through a loop is given by the equation ΔΦ = BAcosθ, where B is the magnetic field strength, A is the area of the loop, and θ is the angle between the normal to the loop and the magnetic field.

In this problem, the magnetic field strength and the area of the loop are given, so we just need to find the change in angle θ as the loop is rotated through half a revolution.

To do this, we can use the fact that the normal to the loop is always parallel to the magnetic field. At the starting position shown in the drawing, the normal is pointing directly to the right (θ = 0). As the loop is rotated through half a revolution, the normal will end up pointing directly to the left (θ = 180°).

Therefore, the change in angle θ is 180°, and the change in magnetic flux is given by ΔΦ = BAcos180° = -BA.

Since the area and magnetic field strength are given in the problem, you can calculate the change in magnetic flux in webers. I hope this helps!