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Q. 83

Expert-verified
Found in: Page 875

### Physics for Scientists and Engineers: A Strategic Approach with Modern Physics

Book edition 4th
Author(s) Randall D. Knight
Pages 1240 pages
ISBN 9780133942651

# Let's look at the details of eddy-current braking. A square CALC loop, lengthon each side, is shot with velocity into a uniform magnetic field . The field is perpendicular to the plane of the loop. The loop has mass and resistance,and it enters the field at. Assume that the loop is moving to the right along the-axis and that the field begins at.a. Find an expression for the loop's velocity as a function of time as it enters the magnetic field. You can ignore gravity, and you can assume that the back edge of the loop has not entered the field.b. Calculate and draw a graph ofover the intervalfor the case that,,,,and . The back edge of the loop does not reach the field during this time interval.

a. The loop velocity as a function of your time is.

b. Velocity over the intervalofis,is,is,is,is and graph is ,

See the step by step solution

## step 1: Calculation for loop velocity (part a)

(a) .

The flux is perpendicular to the loop.

-Magnetic field

-velocity

The loop is subjected to a attractive force generated by

in comparison to the current,

Both sides must be integrated when the speed differs from one side to the opposite.

The field creates the loop.

(b).

## Step 3: Calculation of velocity over the interval part(b) solution

(b).

Because it states,

So,