Question

An electric generator consists of a circular coil of wire of radius 4.0×10−2 m , with 20 turns. The coil is located between the poles of a permanent magnet in a uniform magnetic field, of magnitude 5.0×10−2 T . The B field is orientated perpendicular to the axis of rotation. The ends of the coil are connected via sliding contacts across a resistor of resistance 1.5 Ω. The peak current measured through the resistor is 3.0×10−3 A. What is the angular frequency ω at which the coil is rotating? in rad/s to two sig figs.

Answer 1

Answer:

The angular frequency at which the coil is rotating is 17.9rad/s

Explanation:

To solve the exercise it is necessary to take into account the concepts related to the magnetic field in a wire, Farada's law and Ohm's Law.

The rotational induced voltage is defined by

$V = NBA\omega$

Where,

N = Number of loops

A = Cross-sectional area

$\omega =$angular velocity

B = Magnetic Field

For Ohm's law we have,

V = IR

Where,

I= Current

V = Voltage

R = Resistance

Equation both equations,

$IR = NBA\omega$

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

Our values are given as,

N = 20

$B =5*10^{-2}T$

$R = 1.5\Omega$

$I = 3*10^{-3}A$

$A = \pi r^2= \pi (4*10^{-2})^2$

Replacing the values we have,

$\omega = \frac{(3*10^{-3})(1.5)}{(20)(5*10^{-2})( \pi (4*10^{-2})^2)}$

$\omega = 17.904rad/s$

Therefore the angular frequency at which the coil is rotating is 17.9rad/s