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A copper rod of length *l* is rotated about one end perpendicular to the magnetic field *B* with constant angular velocity ω. The induced e.m.f. between the two ends is

(1) $\frac{1}{2}B\omega {l}^{2}$

(2) $\frac{3}{4}B\omega {l}^{2}$

(3) $B\omega {l}^{2}$

(4) $2B\omega {l}^{2}$

A uniform but time-varying magnetic field *B*(*t*) exists in a circular region of radius *a* and is directed into the plane of the paper, as shown. The magnitude of the induced electric field at point *P* at a distance *r* from the centre of the circular region

(1) Is zero

(2) Decreases as $\frac{1}{r}$

(3) Increases as *r*

(4) Decreases as $\frac{1}{{r}^{2}}$

As shown in the figure, *P* and *Q* are two coaxial conducting loops separated by some distance. When the switch *S* is closed, a clockwise current *I _{P}* flows in

(1) Respectively clockwise and anticlockwise

(2) Both clockwise

(3) Both anticlockwise

(4) Respectively anticlockwise and clockwise

A square metallic wire loop of side 0.1 *m* and resistance of \(1~\Omega\) is moved with a constant velocity in a magnetic field of \(2~\mathrm{wb/m^2}\) as shown in the figure. The magnetic field is perpendicular to the plane of the loop and the loop is connected to a network of resistances. What should be the velocity of the loop so as to have a steady current of 1 *mA* in the loop?

1. | 1 cm/sec |
2. | 2 cm/sec |

3. | 3 cm/sec |
4. | 4 cm/sec |

Shown in the figure is a circular loop of radius *r* and resistance *R*. A variable magnetic field of induction *B* = *B*_{0}*e*^{–t} is established inside the coil. If the key (*K*) is closed, the electrical power developed right after closing the switch, at t=0, is equal to

(1) $\frac{{B}_{0}^{2}\pi {r}^{2}}{R}$

(2) $\frac{{B}_{0}10{r}^{3}}{R}$

(3) $\frac{{B}_{0}^{2}{\pi}^{2}{r}^{4}R}{5}$

(4) $\frac{{B}_{0}^{2}{\pi}^{2}{r}^{4}}{R}$

A rectangular loop with a sliding connector of length *l* = 1.0 *m* is situated in a uniform magnetic field *B* = 2 *T* perpendicular to the plane of the loop. Resistance of connector is *r* = 2 Ω. Two resistances of 6 Ω and 3 Ω are connected as shown in the figure. The external force required to keep the connector moving with a constant velocity *v* = 2 *m/s* is:

1. | 6 N |
2. | 4 N |

3. | 2 N |
4. | 1 N |

A wire *cd* of length *l* and mass *m* is sliding without friction on conducting rails *ax* and *by* as shown. The vertical rails are connected to each other with a resistance *R* between *a* and *b*. A uniform magnetic field *B* is applied perpendicular to the plane *abcd* such that *cd* moves with a constant velocity of

(1) $\frac{mgR}{Bl}$

(2) $\frac{mgR}{{B}^{2}{l}^{2}}$

(3) $\frac{mgR}{{B}^{3}{l}^{3}}$

(4) $\frac{mgR}{{B}^{2}l}$

A conducting rod *AC* of length 4*l* is rotated about a point *O* in a uniform magnetic field $\overrightarrow{B}$ directed into the paper. *AO* = *l* and *OC* = 3*l*. Then

(1) ${V}_{A}-{V}_{O}=\frac{B\omega {l}^{2}}{2}$

(2) ${V}_{O}-{V}_{C}=\frac{7}{2}B\omega {l}^{2}$

(3) ${V}_{A}-{V}_{C}=4B\omega {l}^{2}$

(4) ${V}_{C}-{V}_{O}=\frac{9}{2}B\omega {l}^{2}$

The network shown in the figure is a part of a complete circuit. If at a certain instant the current *i* is 5 *A* and is decreasing at the rate of 10^{3} *A*/*s* then *V _{B}* –

(1) 5

(2) 10

(3) 15

(4) 20

A simple pendulum with bob of mass *m* and conducting wire of length *L* swings under gravity through an angle 2*θ*. The earth’s magnetic field component in the direction perpendicular to swing is *B*. Maximum potential difference induced across the pendulum is

1. $2BL\mathrm{sin}\left(\frac{{\displaystyle \theta}}{{\displaystyle 2}}\right){\left(gL\right)}^{1/2}$

2. $BL\mathrm{sin}\left(\frac{{\displaystyle \theta}}{{\displaystyle 2}}\right){\left(gL\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

3. $BL\mathrm{sin}\left(\frac{{\displaystyle \theta}}{{\displaystyle 2}}\right){\left(gL\right)}^{3/2}$

4. $BL\mathrm{sin}\left(\frac{{\displaystyle \theta}}{{\displaystyle 2}}\right){\left(gL\right)}^{2}$

$\mathrm{An}\mathrm{alternating}\mathrm{current}\mathrm{is}\mathrm{given}\mathrm{as}\mathrm{i}={\mathrm{i}}_{1}\mathrm{cos}\mathrm{\omega t}-{\mathrm{i}}_{2}\mathrm{sin}\mathrm{\omega t}.\mathrm{The}\mathrm{rms}\mathrm{current}\mathrm{is}\mathrm{given}\mathrm{by}$

1. $\frac{{\mathrm{i}}_{1}+{\mathrm{i}}_{2}}{\sqrt{2}}$

2. $\frac{\left|{\mathrm{i}}_{1}+{\mathrm{i}}_{2}\right|}{\sqrt{2}}$

3. $\sqrt{\frac{{\mathrm{i}}_{1}^{2}+{\mathrm{i}}_{2}^{2}}{2}}$

4. $\sqrt{\frac{{\mathrm{i}}_{1}^{2}+{\mathrm{i}}_{2}^{2}}{\sqrt{2}}}$

In a step-up transformer, the turn ratio is 1:2. A Leclanche cell (e.m.f. 1.5V) is connected across the primary coil. The voltage developed in the secondary coil would be-

1. 3.0 V

2. 0.75 V

3. 1.5 V

4. Zero

The peak value of an alternating e.m.f. *E* is given by $E={E}_{0}\mathrm{cos}\omega \text{\hspace{0.17em}}t$ is 10 *volts* and its frequency is 50 *Hz*. At time $t=\frac{1}{600}sec$, the instantaneous e.m.f. is

1. 10 *V*

2. $5\sqrt{3}\text{\hspace{0.17em}}V$

3. 5 *V*

4. 1 *V *

If a current *I* given by ${I}_{0}\mathrm{sin}\text{\hspace{0.17em}}\left(\omega \text{\hspace{0.17em}}t-\frac{{\displaystyle \pi}}{{\displaystyle 2}}\right)$ flows in an ac circuit across which an ac potential of $E={E}_{0}\mathrm{sin}\omega \text{\hspace{0.17em}}t$ has been applied, then the power consumption *P* in the circuit will be

(1) $P=\frac{{E}_{0}{I}_{0}}{\sqrt{2}}$

(2) $P=\sqrt{2}{E}_{0}{I}_{0}$

(3) $P=\frac{{E}_{0}{I}_{0}}{2}$

(4) *P* = 0

A resistance of \(20~ \mathrm{ohms}\) is connected to a source of an alternating potential, \(V=220sin(100 \pi t).\) The time taken by the current to change from its peak value to its r.m.s value will be:

1. | \( 0.2~ \mathrm{sec}\) | 2. | \( 0.25~ \mathrm{sec}\) |

3. | \(25 \times10^{-3}~ \mathrm{sec}\) | 4. | \(2.5 \times10^{-3}~ \mathrm{sec}\) |

In a *LCR* circuit having *L* = 8.0 *henry*, *C* = 0.5 *μF* and *R* = 100 *ohm* in series. The resonance frequency in radian *per second* is** **

(1) 600 *radian/second*

(2) 600 *Hz*

(3) 500 *radian/second*

(4) 500 *Hz*

In a series *LCR* circuit, resistance *R *= 10*Ω* and the impedance *Z *= 20*Ω*. The phase difference between the current and the voltage is

(1) 30°

(2) 45°

(3) 60°

(4) 90°

In the circuit shown below, the ac source has voltage $V=20\mathrm{cos}\left(\omega t\right)$ volts with *ω* = 2000 *rad/sec*.
The amplitude of the current is closest to:

** **

1. 2* *A

2. 3.3* *A

3. $2/\sqrt{5}A$

4. $\sqrt{5}A$

When an ac source of e.m.f. $e={E}_{0}\mathrm{sin}(100$ $t)$ is connected across a circuit, the phase difference between the e.m.f. *e* and the current *i* in the circuit is observed to be \(\frac{\pi}{4}\) as shown in the diagram.
If the circuit consists only of RC or LC in series, then what is the relationship between the two elements?

1. | \(R=1 k \Omega, C=10 \mu F\) |

2. | \(R=1 k \Omega, C=1 \mu F\) |

3. | \(R=1 k \Omega, L=10 \ H\) |

4. | \(R=1 k \Omega, L=1 \ H\) |

In an electrical circuit R, L, C, and an AC voltage source are all connected in series. When L is removed from the circuit, the phase difference between the voltage and the current in the circuit is $\mathrm{\pi}/3.$ If instead, C is removed from the circuit, the phase difference is again $\mathrm{\pi}/3.$ The power factor of the circuit is

(1) 1/2

(2) 1/$\sqrt{2}$

(3) 1

(4) $\sqrt{3}/2$

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