Electromagnetic Induction Test

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Electromagnetic Induction Test - 68

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Weekly Quiz Competition

Question 1
1 / -0

$AB$ is a resistanceless conducting rod which forms a diameter of a conducting ring of radius $r$ rotating in a uniform magnetic field $B$ as shown in figure. The resistors ${R}_{1}$ and ${R}_{2}$ do not rotate. Then the current through the resistor ${R}_{1}$

A
$\cfrac { B\omega { r }^{ 2 } }{ 2{ R }_{ 1 } }$

B
$\cfrac { B\omega { r }^{ 2 } }{ 2{ R }_{ 2 } }$

C
$\cfrac { B\omega { r }^{ 2 } }{ 2{ R }_{ 1 }{ R }_{ 2 } } \left( { R }_{ 1 }{ +R }_{ 2 } \right) \quad$

D
$\cfrac { B\omega { r }^{ 2 } }{ 2\left( { R }_{ 1 }{ +R }_{ 2 } \right) }$

E
answer required

Solution

Emf induced between A and center of circle or B and center of circle

$=\int_0^rB(r\omega)dr=\dfrac{1}{2}B\omega r^2$

This is the also the potential difference across resistor $R_1$ .

Hence, $\dfrac{1}{2}B\omega r^2=i(R_1)$

$\implies i=\dfrac{B\omega r^2}{2R_1}$
Question 2
1 / -0

A vertical conducting ring of radius $R$ falls vertically with a speed $V$ in a horizontal uniform magnetic field which is perpendicular to the plane of the ring. Which of the following statements is correct?

A
$A$ and $B$ are at the same potential

B
$C$ and $D$ are at the same potential

C
current flows in clockwise direction

D
current flows in anticlockwise direction

Solution

$EMF = \int _0 ^{\phi} Bvrd\theta (sin\theta)$
where $\theta =0$ at D
Integrating
$V_{\theta}- V_D= Bvr (1 -cos\theta)$
$V_A = V_D+BvR$
$V_C=V_D$
(Also can be seen by symmetry)
Question 3
1 / -0

A semicircular wire of radius $R$ is rotated with constant angular velocity about an axis passing through one end and perpendicular to the plane of the wire. There is a uniform magnetic field of strength $B$ . The induced emf between the ends is

A
$B\omega {R}^{2}/2$

B
$2B\omega {R}^{2}$

C
is variable

D
none of these

Solution

Induced motional emf in $MNQ$ is equivalent to the motional emf in an imaginary wire $MQ$ .
Therefore,
${ e }_{ MNQ }={ e }_{ MQ }$

$=\dfrac { 1 }{ 2 } { B\omega l }^{ 2 }$

$={ 2B\omega R }^{ 2 }\quad \left(\because l=2R \right)$
Question 4
1 / -0

Directions For Questions

A standing wave $y=2A \sin{kx} \cos{\omega t}$ is set up in the wire $AB$ fixed at both ends by two vertical walls (see figure). The region between the walls contains magnetic field $B$ .

...view full instructions

The wire is found to vibrate in the third harmonic. The maximum emf induced is

A
$\cfrac{4(AB)\omega}{k}$

B
$\cfrac{3(AB)\omega}{k}$

C
$\cfrac{2(AB)\omega}{k}$

D
$\cfrac{(AB)\omega}{k}$

E
answer required

Solution

Since string vibrates in third harmonic, only a length of $\lambda/3$ will produce net emf.

In a segment of length $dx$ , $dE=2A\omega sin(kx)dx$ .

Integrating this in the range $0$ to $\lambda/2$ , we get

$E=\dfrac { 4AB\omega }{ k }$ .
Question 5
1 / -0

Directions For Questions

A metal bar is moving with a velocity of $v=5cm$ ${s}^{-1}$ over a U-shaped conductor. At $t=0$ , the external magnetic field is $0.1T$ directed out of page and is increasing at a rate of $0.2t$ ${s}^{-1}$ . Take $l=5cm$ , and at $t=0,x=5cm$

...view full instructions

The emf induced in the circuit is

A
$125\mu V$

B
$250\mu V$

C
$100\mu V$

D
$300\mu V$

Solution

Area $A=lx$

Induced emf ${ e=\dfrac { d\left( BA \right) }{ dt } }=lx\dfrac { dB }{ dt } +lB\dfrac { dx }{ dt }$

$e={ lx\left( 0.2 \right) }-{ lBv }\quad ...\left( x\quad is\quad decreasing \right)$

At t=0, $e=0.05\times 0.05\times 0.2-0.05\times 0.1\times 0.05=250㎶$
Question 6
1 / -0

A uniform thin rod of length L is moving in a uniform magnetic field $\displaystyle B_{0}$ such that velocity of its centre of mass is v and angular velocity is $\displaystyle \omega=\frac{4v}{L}$ Then

A
e.m.f. between end P and Q of the rod is $\displaystyle B_{0}lv$

B
end P of the rod is at higher potential than end Q of the rod

C
end Q of the rod is at higher potential than end P of the rod

D
the electrostatic field induced in the rod has same direction at all points along the length of rod

Solution

The charges inside the rod move under the magnetic force given by

$\vec{F}=q(\vec{v}\times \vec{B})$

Thus the charges experience force towards end P. Hence end P is at higher potential.

EMF induced along a infinitesimal element along rod of length $dx$ is $dV=Bv(dx)$

$B(x\omega)dx$

Hence emf between points P and Q= $\int_0^L B(x\omega )dx$

$=\dfrac{1}{2}B\omega L^2$

$=2BvL$
Question 7
1 / -0

A thin semicircular conducting ring of radius $R$ is falling with its plane vertical in horizontal magnetic induction $\vec { B }$ . At the position $MNQ$ , the speed of the ring is $V$ , and the potential difference developed across the ring is

A
zero

B
$BV\pi {R}^{2}/2$ and $M$ is at higher potential

C
$\pi RBV$ and $Q$ is at higher potential

D
$2RBV$ and $Q$ is at higher potential

Solution

$E = \cfrac{d \phi}{dt}= B \cfrac{dA}{dt}$

When it is just about to move out:

$dA = 2R (vdt)$

$\cfrac{dA}{dt}= 2 Rv$

Hence, $E = 2BRV$
Question 8
1 / -0

Two coils have a mutual inductance $0.005\ H$ . The current changes in the first coil according to the equation $i = i_{m}\sin \omega t$ where $i_{m} = 10\ A$ and $\omega = 100\pi \ rad\ s^{-1}$ . The maximum value of the emf induced in the second coil is

A
$2\pi$

B
$5\pi$

C
$\pi$

D
$4\pi$

Solution

Given : $L = 0.005 H$
Variation of current in first coil $i = i_m \sin wt$
$\therefore$ $\dfrac{di}{dt} = i_m w \cos wt$
Emf induced $\mathcal{E} = -L\dfrac{di}{dt} = -i_m w L \cos wt$
Thus maximum emf induced $\mathcal{E_{max}} = i_m wL$ when $\cos wt =-1$
$\therefore$ $\mathcal{E_{max}} = 10\times 100\pi \times 0.005 = 5\pi$ volts
Question 9
1 / -0

If emf induced in a coil is $2V$ by changing the current in it from $8\ A$ to $6\ A$ in $2\times 10^{-3}s$ , then the coefficient of self induction is

A
$2\times 10^{-3} H$

B
$10^{-3}H$

C
$0.5\times 10^{-3}H$

D
$4\times 10^{-3}H$

Solution

Induced emf $e = 2V$

$i_{1} = 8A, i_{2} = 6A$

$\triangle t = 2\times 10^{-3}s$

Coefficient of self induction

$L = \dfrac {e}{\triangle i/ \triangle t} = \dfrac {-2}{(6 - 8)/ 2\times 10^{-3}}$

$= \dfrac {-2\times 2\times 10^{-3}}{-2}$

$= 2\times 10^{-3} H$
Question 10
1 / -0

A coil of resistance 5 unit and inductance 4 H is connected to a 10 V battery. The energy stored in thecoil

A
0.8 J

B
8 J

C
16 J

D
4 J

Solution

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Electromagnetic Induction Test - 68

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Question 1

Bωr22R1\cfrac { B\omega { r }^{ 2 } }{ 2{ R }_{ 1 } } 2R1​Bωr2​

Bωr22R2\cfrac { B\omega { r }^{ 2 } }{ 2{ R }_{ 2 } } 2R2​Bωr2​

Bωr22R1R2(R1+R2)\cfrac { B\omega { r }^{ 2 } }{ 2{ R }_{ 1 }{ R }_{ 2 } } \left( { R }_{ 1 }{ +R }_{ 2 } \right) \quad 2R1​R2​Bωr2​(R1​+R2​)

Bωr22(R1+R2)\cfrac { B\omega { r }^{ 2 } }{ 2\left( { R }_{ 1 }{ +R }_{ 2 } \right) } 2(R1​+R2​)Bωr2​

answer required

Solution

Question 2

AAA and BBB are at the same potential

CCC and DDD are at the same potential

current flows in clockwise direction

current flows in anticlockwise direction

Solution

Question 3

BωR2/2B\omega {R}^{2}/2BωR2/2

2BωR22B\omega {R}^{2}2BωR2

is variable

none of these

Solution

Question 4

4(AB)ωk\cfrac{4(AB)\omega}{k}k4(AB)ω​

3(AB)ωk\cfrac{3(AB)\omega}{k}k3(AB)ω​

2(AB)ωk\cfrac{2(AB)\omega}{k}k2(AB)ω​

(AB)ωk\cfrac{(AB)\omega}{k}k(AB)ω​

answer required

Solution

Question 5

125μV125\mu V125μV

250μV250\mu V250μV

100μV100\mu V100μV

300μV300\mu V300μV

Solution

Question 6

e.m.f. between end P and Q of the rod is B0lv\displaystyle B_{0}lvB0​lv

end P of the rod is at higher potential than end Q of the rod

end Q of the rod is at higher potential than end P of the rod

the electrostatic field induced in the rod has same direction at all points along the length of rod

Solution

Question 7

zero

BVπR2/2BV\pi {R}^{2}/2BVπR2/2 and MMM is at higher potential

πRBV\pi RBVπRBV and QQQ is at higher potential

2RBV2RBV2RBV and QQQ is at higher potential

Solution

Question 8

2π2\pi2π

5π5\pi5π

π\piπ

4π4\pi4π

Solution

Question 9

2×10−3H2\times 10^{-3} H2×10−3H

10−3H10^{-3}H10−3H

0.5×10−3H0.5\times 10^{-3}H0.5×10−3H

4×10−3H4\times 10^{-3}H4×10−3H

Solution

Question 10

0.8 J

8 J

16 J

4 J

Solution

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$\cfrac { B\omega { r }^{ 2 } }{ 2{ R }_{ 1 } }$

$\cfrac { B\omega { r }^{ 2 } }{ 2{ R }_{ 2 } }$

$\cfrac { B\omega { r }^{ 2 } }{ 2{ R }_{ 1 }{ R }_{ 2 } } \left( { R }_{ 1 }{ +R }_{ 2 } \right) \quad$

$\cfrac { B\omega { r }^{ 2 } }{ 2\left( { R }_{ 1 }{ +R }_{ 2 } \right) }$

$A$ and $B$ are at the same potential

$C$ and $D$ are at the same potential

$B\omega {R}^{2}/2$

$2B\omega {R}^{2}$

$\cfrac{4(AB)\omega}{k}$

$\cfrac{3(AB)\omega}{k}$

$\cfrac{2(AB)\omega}{k}$

$\cfrac{(AB)\omega}{k}$

$125\mu V$

$250\mu V$

$100\mu V$

$300\mu V$

e.m.f. between end P and Q of the rod is $\displaystyle B_{0}lv$

$BV\pi {R}^{2}/2$ and $M$ is at higher potential

$\pi RBV$ and $Q$ is at higher potential

$2RBV$ and $Q$ is at higher potential

$2\pi$

$5\pi$

$\pi$

$4\pi$

$2\times 10^{-3} H$

$10^{-3}H$

$0.5\times 10^{-3}H$

$4\times 10^{-3}H$