Question

You are looking down on a N = 9 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 cm, and the field goes to zero in t = 0.14 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

Answer:

The magnitude of the voltage is $2.27\times10^{-4}\ V$ and the direction of the current is clockwise.

Explanation:

Given that,

Number of turns = 9

Magnetic field = 0.5 T

Diameter = 3 cm

Time t = 0.14 s

We need to calculate the flux

Using formula of flux

$\phi=NAB$

Put the value into the formula

$\phi=9\times\pi\times(1.5\times10^{-2})^2\times0.5$

$\phi=0.003180$

We need to calculate the emf

Using formula of emf

$\epsilon=-\dfrac{d\phi}{dt}$

$\epsilon=-\dfrac{0.003180}{0.14}$

$\epsilon =-0.000227\ V$

$\epsilon=-2.27\times10^{-4}\ V$

Negative sign shows the direction of current.

Hence, The magnitude of the voltage is $2.27\times10^{-4}\ V$ and the direction of the current is clockwise.