Question

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 . Part A What is the angular frequency ω at which the coil is rotating

Answer 1

Answer:

0.89524 rad/s

Explanation:

N = Number of turns = 20

r = Radius of coil of wire = $4\times 10^{-2}\ m$

B = Magnetic field = $5\times 10^{-2}\ T$

I = Current = $3\times 10^{-3}\ A$

A = Area of coil = $\pi r^2$

R = Resistance = $1.5\ \Omega$

Maximum current in a generator is given by

$I=\frac{NBA\omega}{R}\\\Rightarrow \omega=\frac{IR}{NBA}\\\Rightarrow \omega=\frac{IR}{NB\pi r^2}\\\Rightarrow \omega=\frac{3\times 10^{-3}\times 1.5}{20\times 5\times 10^{-2}\times \pi\times (4\times 10^{-2})^2}\\\Rightarrow \omega=0.89524\ rad/s$

The angular frequency is 0.89524 rad/s

Answer 2

Answer:

w = 0.8957 rad/sec

Explanation:

we know that Maximum current in coil can be calculated as

$I = \frac{NBAw}{R}$

where N represent number of turn = 20

B = magnetic field $=  5 \times 10^{-2} T$

R is resistance  =  1.5 ohm

$I = 3.0\times 10^{-3} A$

A = area $= \pi r^2$

solving for angular frequency w

$w  = \frac{I R}{NB \pi r^2}$

$w = \frac{3.0\times 10^{-3} \times 1.5}{20\times 5\times 10^{-2} \pi \times(4\times 10^{-2})^2}$

w = 0.8957 rad/sec