Electromagnetic Induction Test

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Electromagnetic Induction Test - 63

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Question 1
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Electromagnetic induction is

A
Charging a substance.

B
Process of developing a magnetic field around a coil by passing electricity through a coil

C
Process of rotating the armature of a generator

D
Process of making electricity by the relative motion of a magnet or a coiled conductor.

Solution

Hint: The creation of an electromotive force across an electrical conductor in a changing magnetic field is known as electromagnetic induction.

Correct answer: Option D

Explanation of correct answer:
When a coil is moved close to a magnet and a relative motion between the two is generated by moving either the magnet or the coil, the magnetic flux connections via the coil change. An emf and, as a result, an electric current flow through the coil as a result of the change in magnetic flux.
Therefore, the correct answer is option (D).

Explanation of Incorrect Options:
During the charging by induction procedure, the charged particle is held near an uncharged conductive substance grounded on a neutrally charged material. There is no relative motion between the magnet and the coiled conductor. So, Option (A) is incorrect.
current must flow when a changing magnetic field passes through a wire coil. Due to the fact that the magnetic field remains constant. So, Option (B) is incorrect.
The process of rotating the armature of a generator counteracts the induction cause.. So, Option (c) is incorrect.

Therefore, option (A), option (B) and option (C) are incorrect. The correct answer is option (D).
Question 2
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A conducting rod of length l moves with velocity v in x-direction axis parallel to a long wire carrying a steady current I. The axis of the rod is maintained perpendicular to the wire with near end a distance r away as shown in the fig. Find the emf induced in the rod.

A
$$\dfrac{\mu _{0}I\upsilon }{\pi }\, \, ln \left ( \frac{r+1}{r} \right )$$

B
$$\dfrac{2\mu _{0}I\upsilon }{\pi }\, \, ln \left ( \frac{r+1}{r} \right )$$

C
$$\dfrac{\mu _{0}I\upsilon }{\pi }\, \, ln \left ( \dfrac{l}{r+1} \right )$$

D
$$\dfrac{\mu _{0}I\upsilon }{ 2 \pi }\, \, ln \left ( \dfrac{r+1}{r} \right )$$

Solution
Question 3
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A rectangular loop has a sliding connector PQ of length $$l$$ and resistance $$\mathrm{R}\Omega$$ and it is moving with a speed $$\mathrm{v}$$ as shown. The set-up is placed in a uniform magnetic field going into the plane of the paper. The three currents $$I_{1},\ I_{2}$$ and $$I$$ are

A
$$I_{1}=-I_{2}=\displaystyle \frac{\mathrm{B}l\mathrm{v}}{\mathrm{R}},\ I=\displaystyle \frac{2\mathrm{B}l\mathrm{v}}{\mathrm{R}}$$

B
$$I_{1}=I_{2}=\displaystyle \frac{\mathrm{B}l \mathrm{v}}{3\mathrm{R}},\ I=\displaystyle \frac{2\mathrm{B}l \mathrm{v}}{3\mathrm{R}}$$

C
$$I_{1}=I_{2}=I=\displaystyle \frac{\mathrm{B}l\mathrm{v}}{\mathrm{R}}$$

D
$$I_{1}=I_{2}=\displaystyle \frac{\mathrm{B}l \mathrm{v}}{6\mathrm{R}},\ I=\displaystyle \frac{\mathrm{B}l \mathrm{v}}{3\mathrm{R}}$$

Solution

A moving conductor is equivalent to a battery of emf $$=\mathrm{v}\mathrm{B}\ell$$ (motion emf) Equivalent circuit

$$I=I_1+I_2$$

Applying Kirchoff's law
$$I_{1}\mathrm{R}+I\mathrm{R}-\mathrm{v}\mathrm{B}l=0$$ ............... (1)
$$I_{2}\mathrm{R}+I \mathrm{R}-\mathrm{v} \mathrm{B}P=0$$ ............... (2)

Adding $$(1)$$ and $$(2)$$
$$2I\mathrm{R}+I\mathrm{R}=2\mathrm{v}\mathrm{B}l $$
$$\displaystyle I=\frac{2\mathrm{v}\mathrm{B}l}{3\mathrm{R}} $$

Solving, we get
$$I_{1}=I_{2}=\displaystyle \frac{\mathrm{v}\mathrm{B}\ell}{3\mathrm{R}}$$
Question 4
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In a uniform magnetic field of induction $$B$$, a wire in the form of semicircle of radius $$r$$ rotates about the diameter of the circle with angular velocity $$\omega$$. If the total resistance of the circuit is $$R$$, the mean power generated per period of rotation is :

A
$$\dfrac{B\pi r^{2}\omega }{2R}$$

B
$$\dfrac{(B\pi r^{2}\omega)^{2} }{8R}$$

C
$$\dfrac{(B\pi r\omega)^{2} }{2R}$$

D
$$\dfrac{(B\pi r\omega^{2})^{2} }{8R}$$

Solution

Flux $$\phi = BA\cos \theta=B \times ( \dfrac{1}{2}\pi r^2) \cos \theta $$
$$| \varepsilon| =| \dfrac{ d \phi}{dt} |= \dfrac{1}{2}\pi Br^2 \dfrac{d(\cos \theta)}{dt}$$
$$| \varepsilon|= \dfrac{1}{2}\pi Br^2( \sin \theta) \dfrac{d\theta}{dt}$$
$$| \varepsilon| = \dfrac{1}{2}\pi Br^2( \sin \omega t)\omega $$
Power: $$P = \dfrac{| \varepsilon| ^2}{R}= \dfrac{(\dfrac{1}{2}\pi Br^2\omega \sin \omega t )^2}{R}$$
Mean Power $$P_{mean} = \dfrac{ \int_{0}^{T} \dfrac{| \varepsilon| ^2}{R}dt} {\int_{0}^{T}dt}= \dfrac{\pi ^2 B^2r^4\omega ^2 \int_{0}^{T} \sin ^2\omega t\: dt }{4TR}$$
$$= \dfrac{\pi ^2 B^2r^4\omega ^2}{4TR}\frac{T}{2}= \dfrac{\pi ^2 B^2r^4\omega ^2}{8R}$$
Question 5
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A straight conducting rod of length $$30$$ cm and having a resistance of $$0.2$$ ohm is allowed to slide over two parallel thick metallic rails with uniform velocity of $$0.2$$ m/s as shown in the figure. The rails are situated in a horizontal plane. If the horizontal component of earth's magnetic field is $$0.3\times 10 ^{-4}$$T and a steady state current of $$3\mu A$$ is induced through the rod. The angle of dip will be :

A
$$tan^{-1}\left ( \dfrac{3}{4} \right )$$

B
$$tan^{-1}\left ( \dfrac{1}{\sqrt{3}} \right )$$

C
$$tan^{-1}(\sqrt{3})$$

D
$$tan^{-1}\left ( \dfrac{1}{3} \right )$$

Solution

Only vertical lines of earth's magnetic field will cut when the rod is allowed to slide over the rails.
Let the vertical compents of the earth's magnetic field is $$ B_v $$.
Induced current $$ i=\dfrac{B_v l v}{R} $$
$$3 \times 10^{-6} = \dfrac{B_v \times 0.3 \times 0.2}{0.2}$$
$$ B_v =10^{-5} T$$
given $$ B_H =0.3\times 10^{-4} T $$
The angle of dip $$ \theta $$ is given by $$ = tan^{-1} ( \dfrac{B_v}{B_H} ) = tan^{-1}(\dfrac{1}{3}) $$
Question 6
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Directions For Questions

A metallic rod ab of length 1.5 m and of uniform cross-section with a resistance of 2.5 $$\Omega$$ is inclined to +ve X-axis at an angle of 30$$^{o}$$. It moves with a speed of 3 ms$$^{-1}$$ along +ve X-axis through a uniform magnetic field of 2 T, oriented along +ve Z-axis as shown in the Figure. An ammeter A of 0.5 $$\Omega$$ resistance is connected across ab.

...view full instructions

The induced emf developed across the rod must be

A
4.5 V with b at higher potential

B
1.5 V with a at higher potential

C
1.5 V with b at higher potential

D
4.5 V with a at higher potential

Solution

The emf induced is
$$=B\cdot \overrightarrow{l}\times \overrightarrow{V}$$
$$=(2)\times 1.5\times 3\times Sin 30^{\circ}$$
$$=4.5 V$$ with at higher potential.
Question 7
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Directions For Questions

Two parallel vertical metallic rails AB and CD are separated by $$1m$$. They are connected at the ends by resistances $$\mathrm{R}_{1}$$ and $$\mathrm{R}_{2}$$ as shown in figure. A horizontal metallic bar of length $$L$$ and mass $$0.2 kg$$ slides without friction vertically down the rails under tbe action of gravity. There is a uniform horizontal magnetic field of strength $$\mathrm{B}=0.6$$ tesla perpendicular to the plane of the rails. lt is observed that when the terminal velocity is attained, the power dissipated in $$\mathrm{R}_{\mathrm{1}}$$ and $$\mathrm{R}_{2}$$ are $$\mathrm{P}_{1}$$ $$=$$ $$0.76$$ watts and $$\mathrm{P}_{2}$$ $$=$$ $$1.2$$ watts respectively.

...view full instructions

What is lts teminal velocity?

A
$$1.5 m/sec$$

B
$$0.8 m/sec$$

C
$$0.7 m/sec$$

D
$$1 m/sec$$

Solution

The rod becomes a seat of e.m.f $$e=Blv$$ when it accquire the terminal velocity the resultant force on it is zero
$$mg-Bil=0$$ or $$mg=Bil$$
The equivalent corcuit is shown

$$i=\dfrac{e}{\dfrac{R_{1}R_{2}}{R_{1}+R_{2}}}$$

$$=\dfrac{Blv(R_{1}+R_{2})}{R_{1}R_{2}}$$

$$mg=\dfrac{BBlv(R_{1}+ R_{2})}{R_{1}R_{2}} \Rightarrow v=\dfrac{mgR}{B^2I^2}$$

$$e=Blv=\dfrac{mgR}{Bl}$$

$$P_{1}=\dfrac{e^{2}}{R_{1}}=\dfrac{m^{2}g^{2}R^{2}}{B^{2}l^{2}R_{1}}=\dfrac{m^{2}g^{2}}{B^{2}l^{2}R_{1}}\times \dfrac{R_{1}^{2}R_{2}^{2}}{(R_{1}+R_{2})^{2}}$$

$$\Rightarrow \dfrac{(R_{1}+R_{2})^{2}}{R_{1}R_{2}^{2}}=\dfrac{m^{2}g^{2}}{B^{2}l^{2}P_{1}}$$

$$\Rightarrow \dfrac{1}{R_{1}}\dfrac{(R_{1}+R_{2})^{2}}{R_{2}}=\dfrac{m^{2}g^{2}}{B^{2}l^{2}P_{1}}$$

Similarly,$$\dfrac{1}{R_{2}}\left ( \dfrac{R_{1}+R_{2}}{R_{1}} \right )^{2}=\dfrac{m^{2}g^{2}}{B^{2}l^{2}P_{2}}$$

Dividing$$\dfrac{R_{2}}{R_{1}}=\dfrac{P_{1}}{P_{2}}\Rightarrow \dfrac{R_{2}+R_{1}}{R_{2}}=\dfrac{P_{1}+P_{2}}{P_{1}}$$

$$\dfrac{1}{R_{1}}\left ( \dfrac{P_{1}+P_{2}}{P_{1}} \right )^{2}=\dfrac{m^{2}g^{2}}{B^{2}l^{2}P_{1}}\Rightarrow R_{1}=\dfrac{B^{2}l^{2}}{m^{2}g^{2}}\dfrac{(P_{1}+P_{2})^{2}}{P_{1}}$$

$$R_{2}=\dfrac{b^{2}l^{2}}{m^{2}g^{2}}\dfrac{(P_{1}+P_{2})^{2}}{P_{2}}=\dfrac{(0.76+1.2)^{2}}{1.2}\times \dfrac{0.6^{2\times 1^{2}}}{0.2^{2}\times 9.8^{2}}$$

$$=0.3ohm$$

$$R_{1}=\dfrac{(0.76+1.2)^{2}}{0.76}\times \dfrac{0.6^{2}\times 1^{2}}{0.2^{2}\times 9.8^{2}}=0.47ohm$$

$$v=\dfrac{mg}{B^{2}l^{2}}.\dfrac{R_{1}R_{2}}{(R_{1}+R_{2})}=1m/sec$$
Question 8
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A square metal wire loop of side 10 cm and of resistance 2 $$\Omega $$ moves with constant velocity in the presence of a uniform magnetic field of induction 4 T, perpendicular and into the plane of the loop. The loop is connected to a network of resistance as shown in the Figure. If the loop should have a steady current of 2 mA, the speed of the loop must be (in cm s$$^{-1}$$) :

A
$$4$$

B
$$2$$

C
$$3$$

D
$$6$$

Solution

As this is wheatstore bridge.

$$R_{eq}=\dfrac{6\times3}{6+3}=2\Omega$$

$$Total resistance=2+2=4\Omega$$
Now steady current is 2mA
$$\therefore $$ Voltage required $$=(.002)4$$
$$=0.008V=8mV$$
$$B=4T$$

$$E=8mV=\dfrac{-d\phi}{dt}$$$$=\dfrac{-d4(Area)}{dt}$$

$$8\times10^{-3}=4\times(0.10)V$$
$$V=2\times 10^{-2}m/s$$
$$=2cm/s$$
Question 9
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A conductor rod AB moves parallel to X- axis in a uniform magnetic field, pointing in the positive Z-direction. The end A of the rod gets-

A
positively charged

B
negatively charge

C
neutral

D
first positively charged and then negatively chrged

Solution

We know that the force on a charge is given by $$F=q\vec{v}\times \vec{B}$$

Here, on a positive charge, $$\hat{v}=\hat{i}$$ and $$\hat{B}=\hat{k}$$

Hence, force on a positive charge is along $$\hat {i}\times \hat {k}=-\hat{j}$$

Hence, end B got positively charged and end A get negatively charged.

Answer-(B).
Question 10
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There is a uniform magnetic field B normal to the xy plane. A conductor ABC has length $$AB=l_{1}$$, parallel to the x-axis, the length $$BC=l_{2}$$, parallel to the y axis. ABC moves in the xy plane with velocity $$ v_{x} \hat {i} + v_{y} \hat{j} $$. The potential difference between A and C is proportional to :

A
$$v_{x}l_{1} + v_{y} l_{2}$$

B
$$v_{x}l_{2} + v_{y} l_{1}$$

C
$$v_{x}l_{2} - v_{y} l_{1}$$

D
$$v_{x}l_{1} - v_{y} l_{2}$$

Solution

Let magnetic field be $$B\hat{k}$$
$$\vec{v}=v_x\hat{i} +v_y\hat{j}, \\
V_B-V_A=(l_1 \hat{i}).(\vec{v} \times B \hat{k})= B v_y l_1 \\
V_C-V_B=(l_2 \hat{j}).(\vec{v} \times B \hat{k})= -B v_x l_2 \\
\Rightarrow V_A-V_C=B v_x l_2-B v_y l_1
$$