# Thread: change in magnetic flux

1. ## change in magnetic flux

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

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.

2. Originally Posted by rcmango
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

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.
$\displaystyle \Phi _m = \vec{B} \cdot \vec{A}$
So find the angle that A makes with B at time t and take the dot product.

A0 = x0*L
what is this and where did it come from?

-Dan

3. okay, i haven't learned dot product in for calc yet, this physics class isn't calculus based. Just checking to see if there is another way to do this one instead of dot product?

thankyou so far.

4. Originally Posted by rcmango
okay, i haven't learned dot product in for calc yet, this physics class isn't calculus based. Just checking to see if there is another way to do this one instead of dot product?

thankyou so far.
$\displaystyle \Phi _m = \vec{B} \cdot \vec{A} = BA~cos(\theta)$
where $\displaystyle \theta$ is the angle between B and A.

-Dan