Question

A circular coil of radius $8.0\,cm$ and $20$ turns is rotated about its vertical diameter with an angular speed of $50\,rad\,{s^{ - 1}}$ in a uniform horizontal magnetic field of magnitude $3.0 \times {10^{ - 2}}\,T$ . Obtain the maximum and average emf induced in the coil. If the coil forms a closed loop of resistance $10\,\Omega$ , calculate the maximum value of current in the coil. Calculate the average power loss due to Joule heating. Where does this power come from?

Answer 1

The induced emf of a coil placed in a magnetic field is given by $e = NAB\omega$ where $e$ is the induced emf, $N$ is the number of turns, $A$ is the area of cross section, $B$ is the magnetic field and $\omega$ is the angular frequency. Having known the induced emf and the resistance of a coil, we can easily find out the maximum current in the coil using the relation $i = \dfrac{e}{R}$ where $i$ is the current and $R$ is the resistance. The average power can be simply calculated by using the formula $P = \dfrac{{ei}}{2}$ . We need to first convert all the quantities known to us in their SI units and then apply them in the respective formula.

Complete step by step answer:
Given that $\omega = 50\,rad\,{s^{ - 1}}$ , $N = 20$ , $B = 3.0 \times {10^{ - 2}}\,T$.
We know that the radius of the coil is $r = 8.0\,cm$ .
Converting to SI units
$r = 0.08\,m$
So, the area of cross section becomes $A = \pi {r^2}$
$A = 3.14 \times 0.0064$
$\Rightarrow A \approx 0.02\,{m^2}$
We know that $e = NAB\omega$
Substituting the values, we get
$e \approx 20 \times 0.02 \times 0.03 \times 50$
$\Rightarrow e \approx 0.600\,V$

Now given that the resistance of the coil is $R = 10\,\Omega$ and we know that $i = \dfrac{e}{R}$
So, we can substitute the values in the equation to get,
$i \approx \dfrac{{0.600}}{{10}}$
$\Rightarrow i \approx 0.06\,A$
Now the average power can be calculated as $P = \dfrac{{ei}}{2}$
Substituting the values,
$P = \dfrac{{0.6 \times 0.06}}{2}$
$\therefore P = 0.018\,J{s^{ - 1}}$

Therefore, $0.018\,J{s^{ - 1}}$ power comes from the heating of the rod in motion when placed in a magnetic field.

Note: The average power is calculated over a cycle. We can also calculate instantaneous power and then integrate it over one full cycle to get the result. But since the magnetic field was uniform, we simply applied the formula. But if the magnetic field was varying or the change in the magnetic flux was not constant, we would have used the integration approach to solve the question.