Q1GATE 2016NAT

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 __________.

🚀 Solve in Practice Mode

📖 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}

Q2GATE 2015NAT

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 ________.

🚀 Solve in Practice Mode

📖 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}

Q3GATE 2015NAT

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 _________.

🚀 Solve in Practice Mode

📖 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}

Time Varying Fields Practice Questions

Get the full Time Varying Fields experience — 100% free