Question

You are looking down on a N = 17 turn coil in a magnetic field B = 0.5 T which points directly down into the screen. If the diameter of the coil d = 3.8 cm, and the field goes to zero in t = 0.24 seconds, what would be the magnitude of the voltage (in Volts) and direction of the induced current? Indicate the direction of the current by the sign in front of your voltage: counterclockwise is positive, clockwise is negative.

Answer 1

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

The voltage is
`\epsilon = 0.40163 \ V`

The direction of the induced current is clockwise

Step-by-step explanation:

From the question we are told that

The number of turns is N = 17

The magnetic field is
`B_2 = 0.5 \ T`

The diameter is
`d = 3.8 \ cm = 0.038 \ m`

The time interval is
`dt = 0.24 \ s`

The induce emf is mathematically represented as

`\epsilon = - N (d\phi)/(dt)`

`\epsilon = - N (d ( B_2 - B_1 ) A )/(dt)`

Here
`B_1` is the magnetic field experienced by the coil before entering the magnetic field given in the question i.e
`B_1 = 0`

Here the negative sign show that the induced voltage is moving in a direction opposite to the change magnetic flux

The area is mathematically represented as

`A = \pi (d^2)/(4)`

=>
`A = 3.142 * ( 0.038^2 )/(4)`

=>
`A = 0.01134 \ m^2`

Hence

`\epsilon = - 17 * ( 0.5 * 0.01134 )/( 0.24)`

`\epsilon = 0.40163 \ V`

The direction of the induced current is the same as that of induced voltage

Thus the direction is clockwise