Question 1 |
A rotating conductor of 1 m length is placed in a radially outward (about the z-axis) magnetic flux
density (B) of 1 Tesla as shown in figure below. Conductor is parallel to and at 1 m distance from the z-axis. The speed of the conductor in r.p.m. required to induce a voltage of 1 V across it, should
be __________.
2.62 | |
8.34 | |
7.75 | |
9.55 |
Question 1 Explanation:
Voltage insuced =\int_{0}^{1}E_m \cdot dt
(where E_m is induced electric field)
Since,
Volatage induced =1 V
So, E_m = 1 V/m
As we know, E_m=\bar{V} \times \bar{B}
where, V= (radius of path) x (angular velocity)
\begin{aligned} \frac{1V}{m} &=(V \times 1 \; \text{Tesla}) \\ V&=1 \; m/sec \\ V&= r \times \omega =1 \; m/sec\\ \text{Since, } r&=1m, \; \text{So }\omega =1 \; \text{rad/sec} \end{aligned}
Now from this we get
\begin{aligned} \omega &=2 \times \pi \times \frac{N}{60}=1 \; \text{rad/sec}\\ N&=\frac{30}{\pi}=9.55 \; \text{revolutions per minute} \end{aligned}
(where E_m is induced electric field)
Since,
Volatage induced =1 V
So, E_m = 1 V/m
As we know, E_m=\bar{V} \times \bar{B}
where, V= (radius of path) x (angular velocity)
\begin{aligned} \frac{1V}{m} &=(V \times 1 \; \text{Tesla}) \\ V&=1 \; m/sec \\ V&= r \times \omega =1 \; m/sec\\ \text{Since, } r&=1m, \; \text{So }\omega =1 \; \text{rad/sec} \end{aligned}
Now from this we get
\begin{aligned} \omega &=2 \times \pi \times \frac{N}{60}=1 \; \text{rad/sec}\\ N&=\frac{30}{\pi}=9.55 \; \text{revolutions per minute} \end{aligned}
Question 2 |
A circular turn of radius 1 m revolves at 60 rpm about its diameter aligned with the x-axis as shown in the figure. The value of \mu _{0} is 4\pi \times 10^{-7} in SI unit. If a uniform magnetic field intensity \vec{H}=10^{7}\hat{z} A/m is applied, then the peak value of the induced voltage, V_{turn} ( in Volts), is _________.
150.35 | |
200.25 | |
248.05 | |
300.54 |
Question 2 Explanation:
The circular turn rotate with 60 rpm, let the angle made by ring w.r.t. x-axis \theta
and \;\;\; \theta =\omega _0 t
the turn rotate at 60 rpm,
so, \;\;\; \omega _0=2 \pi
So, the flux flowing through the circular turn wil be
\begin{aligned} \Psi &=(\mu _0 H_z \times \text{Area of turn} \times \cos \omega _0 t) \\ \Psi &= 4 \pi \times 10^{-7} \times 10^7\; A/m \times \pi \times 1^2 \times \cos \omega _0 t\\ &\text{Maximum voltage induced is } \\ \\ \left. \begin{matrix} \frac{d\Psi}{dt} \end{matrix}\right|_{max}&=(\omega _0 \times 4 \pi \times \pi \sin \omega _0 t)_{max}\\ V_{max}&=(4 \pi^2 \times 2 \pi)=248.05 volts \end{aligned}
Question 3 |
Consider a one-turn rectangular loop of wire placed in a uniform magnetic field as shown in the figure. The plane of the loop is perpendicular to the field lines. The resistance of the loop is 0.4 \Omega, and its inductance is negligible. The magnetic flux density (in Tesla) is a function of time, and is given by B(t)=0.25 \sin \omega t, where \omega =2\pi \times 50 radian/second. The power absorbed (in Watt) by the loop from the magnetic field is ________.
0.11 | |
0.34 | |
0.96 | |
0.19 |
Question 3 Explanation:
\begin{aligned} A&=10cm \times 5 cm\\ B(t)&=0.25 \sin \omega t,\; R=0.4\Omega \\ P&=i^2 R\\ \because \; e&=-\frac{d\phi }{dt}=-\frac{d}{dt}(B \times A)\\ &=-\frac{d}{dt}[0.25 \times 50 \times 10^{-4} \sin \omega t]\\ e&=-0.25 \times 50 \times 10^4 \omega \cos \omega t\\ &=-1.25 \times 10^{-3}\omega \cos \omega t\\ P_{evg}&=\frac{e^2_{rms}}{R}\\ P&=\int_{0}^{T}\frac{(12.5 \times 100 \pi \times 10^{-4})^2}{0.4} \times \cos ^2 \omega t d(\omega t)\\ &=(12.5 \times \pi \times 10^{-2})^2 \times \frac{1}{2 \times 0.4}\\ &=0.193\; \text{Watt} \end{aligned}
There are 3 questions to complete.