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.

Question

contestada

Respuesta :

CarliReifsteck CarliReifsteck · Answer 1 · 2019-08-08T13:18:24+02:00

Answer:

The magnitude of the voltage is [tex]2.27\times10^{-4}\ V[/tex] 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

[tex]\phi=NAB[/tex]

Put the value into the formula

[tex]\phi=9\times\pi\times(1.5\times10^{-2})^2\times0.5[/tex]

[tex]\phi=0.003180[/tex]

We need to calculate the emf

Using formula of emf

[tex]\epsilon=-\dfrac{d\phi}{dt}[/tex]

[tex]\epsilon=-\dfrac{0.003180}{0.14}[/tex]

[tex]\epsilon =-0.000227\ V[/tex]

[tex]\epsilon=-2.27\times10^{-4}\ V[/tex]

Negative sign shows the direction of current.

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