Question

A semicircular wire of radius $r$ is rotating in transverse magnetic field $B$ with angular frequency $\omega$ about the diameter. A resistor of resistance $R$ is connected with the help of conducting brushes as shown. Ignore the other resistance and magnetic effect of induced current. The average thermal power generated in the resistor is $\frac{(\pi B\omega)^2r^n}{mR}$. Find $\frac{m}{n}$

Answer 1

2

Answer 2

The problem asks us to find the ratio $\frac{m}{n}$ based on the formula for the average thermal power generated in a resistor connected to a rotating semicircular wire in a uniform magnetic field.

1. Magnetic Flux:
   The area of the semicircular loop is $A = \frac{1}{2}\pi r^2$.
   As the semicircle rotates about its diameter in a uniform transverse magnetic field $B$ with angular frequency $\omega$, the magnetic flux $\Phi$ through the loop varies with time. Assuming the plane of the semicircle is perpendicular to the magnetic field at $t=0$, the magnetic flux at any time $t$ is given by: $\Phi(t) = B A \cos(\omega t)$ Substituting the area $A$: $\Phi(t) = B \left(\frac{1}{2}\pi r^2\right) \cos(\omega t)$

2. Induced Electromotive Force (EMF):
   According to Faraday's law of electromagnetic induction, the induced EMF $\varepsilon$ is the negative rate of change of magnetic flux: $\varepsilon = -\frac{d\Phi}{dt}$ $\varepsilon = -\frac{d}{dt} \left[ B \left(\frac{1}{2}\pi r^2\right) \cos(\omega t) \right]$ $\varepsilon = -B \left(\frac{1}{2}\pi r^2\right) (-\omega \sin(\omega t))$ $\varepsilon = \frac{1}{2} B \pi r^2 \omega \sin(\omega t)$

3. Instantaneous Power:
   The instantaneous current $I(t)$ flowing through the resistor $R$ is $I(t) = \frac{\varepsilon}{R}$.
   The instantaneous thermal power $P(t)$ generated in the resistor is given by $P(t) = I(t)^2 R = \frac{\varepsilon^2}{R}$: $P(t) = \frac{\left(\frac{1}{2} B \pi r^2 \omega \sin(\omega t)\right)^2}{R}$ $P(t) = \frac{\frac{1}{4} B^2 \pi^2 r^4 \omega^2 \sin^2(\omega t)}{R}$

4. Average Thermal Power:
   To find the average thermal power $\langle P \rangle$, we average $P(t)$ over one complete cycle. The average value of $\sin^2(\omega t)$ over a complete cycle is $\frac{1}{2}$. $\langle P \rangle = \frac{\frac{1}{4} B^2 \pi^2 r^4 \omega^2}{R} \langle \sin^2(\omega t) \rangle$ $\langle P \rangle = \frac{B^2 \pi^2 r^4 \omega^2}{4R} \left(\frac{1}{2}\right)$ $\langle P \rangle = \frac{B^2 \pi^2 r^4 \omega^2}{8R}$

5. Comparing with the Given Form:
   The derived average thermal power is $\langle P \rangle = \frac{B^2 \pi^2 r^4 \omega^2}{8R}$.
   We can rewrite this as: $\langle P \rangle = \frac{(\pi B\omega)^2 r^4}{8R}$ The problem states that the average thermal power is $\frac{(\pi B\omega)^2r^n}{mR}$.
   Comparing the two expressions: $n = 4$ $m = 8$

6. Finding $\frac{m}{n}$:
   Finally, we calculate the ratio $\frac{m}{n}$: $\frac{m}{n} = \frac{8}{4} = 2$

The final answer is $\boxed{2}$.

Explanation of the solution: The magnetic flux through the rotating semicircle is $\Phi = B A \cos(\omega t)$, where $A = \frac{1}{2}\pi r^2$. The induced EMF is $\varepsilon = -\frac{d\Phi}{dt} = \frac{1}{2} B \pi r^2 \omega \sin(\omega t)$. The instantaneous power is $P = \frac{\varepsilon^2}{R} = \frac{B^2 \pi^2 r^4 \omega^2 \sin^2(\omega t)}{4R}$. The average power is $\langle P \rangle = \frac{B^2 \pi^2 r^4 \omega^2}{4R} \langle \sin^2(\omega t) \rangle = \frac{B^2 \pi^2 r^4 \omega^2}{4R} \left(\frac{1}{2}\right) = \frac{(\pi B\omega)^2 r^4}{8R}$. Comparing this with the given form $\frac{(\pi B\omega)^2r^n}{mR}$, we find $n=4$ and $m=8$. Therefore, $\frac{m}{n} = \frac{8}{4} = 2$.