Question

A flat, square coil of 18 turns that has sides of length 14.0 cm is rotating in a magnetic field of strength 0.040 T. If the maximum emf produced in the coil is 36.0 mV, what is the angular velocity of the coil (in rad/s)? (Enter the magnitude.)

Answer 1

The angular velocity of the coil is
`\omega =2.55 rad s^(-1)`.

Step-by-step explanation:

The expression for the maximum emf is as follows;

`\epsilon =NBA\omega` ......... (1)

Here,
`\epsilon` is the emf, {tex]\omega[/tex] is the angular velocity, N is the number of turns, B is the magnetic field and A is the area.

Calculate the area of the square coil.

Convert side of the length form cm to m.

`s=(14)/(100)m`

s= 0.14 m

`A=s^(2)`

Here, A is the area and s is the length of the side of the square.

Put s= 0.14 m.

`A=0.0196 m^(2)`

Convert maximum emf from mV to V.

`\epsilon =36* 10^(-3) V`

Calculate the angular velocity of the coil by rearranging the equation (1).

`\omega =(\epsilon )/(NBA)`

Put
`A=0.0196 m^(2)`,
`\epsilon =36* 10^(-3) V`, B= 0.040 T and N= 18 turns.

`\omega =(36* 10^(-3) )/(18(0.040)(0.0196))`

`\omega =2.55 rad s^(-1)`

Therefore, the angular velocity of the coil is
`\omega =2.55 rad s^(-1)`.