Question

B Crank In the figure provided, a magnet creates a magnetic field of strength 3.63 T. A hand crank generator consists of a semi circle with a radius of r = 5.50 cm as shown in the figure. The resistance in the bulb is 7.20 9. With what frequency will you have to turn the crank for the bulb to dissipate a power that has a maximum value of 6.00 W? Answer: 363 rad/s

Answer 1

The periodic e.m.f induced due to rotation of a coil in a uniform magnetic field is given by

[where A = area , N = number of turns, B = magnetic field, f = frequency]

E = 27 N BAf sin (20 ft)

The term  $2\pi NBAf$ is maximum induced e.m.f.

Now, in our problem:

N = 1

B = 3.63 T

r = 5.5 cm
A = $\frac{\pi r^{2}}{2}$ [for semicircle]

R = resistance of bulb = 7.2 ohm

P = maximum power dissipated = 6 W

For maximum power to dissipate, $P_{max} = \frac{E.M.F^{2}_{max}}{R}$

$\Rightarrow E.M.F_{max}=\sqrt{RP_{max}}$

$\Rightarrow 2\pi*B*\frac{\pi*r^{2}}{2}*f=\sqrt{RP_{max}}$

$\Rightarrow f=\frac{\sqrt{RP_{max}}}{2\pi*B*\frac{\pi*r^{2}}{2}}$

$=\frac{\sqrt{7.2*6}}{\pi*3.63*\pi*(5.5*10^{-2})^{2}}=60.65 Hz$ [answer]