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

Answer:

The angular velocity of the coil is $\omega =2.55 rad s^{-1}$.

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.

$s=\frac{14}{100}m$

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.

$\epsilon =36\times 10^{-3} V$

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

$\omega =\frac{\epsilon }{NBA}$

Put $A=0.0196 m^{2}$, $\epsilon =36\times 10^{-3} V$, B= 0.040 T and N= 18 turns.

$\omega =\frac{36\times 10^{-3} }{18(0.040)(0.0196)}$

$\omega =2.55 rad s^{-1}$

Therefore, the angular velocity of the coil is $\omega =2.55 rad s^{-1}$.