Electromagnetic Induction Test

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Electromagnetic Induction Test - 28

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Question 1
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A thin wire of length $$2$$$$m$$ is perpendicular to the $$x-y$$ plane. It is moved with velocity $$\vec { v } = (2\hat { i } + 3 \hat { j } + \hat { k }) m/s$$ through a region of magnetic induction $$\vec {B} = (\hat {i} + 2\hat{j}) Wb/m^{2}.$$ Then potential difference induced between the ends of the wire is

A
$$2V$$

B
$$4V$$

C
$$0 V$$

D
none of these

Solution

Induced emf $$\displaystyle \varepsilon = \int_{0}^{l}( \vec{v} \times \vec{B}) .\vec{dl}= \int_{0}^{l}( \vec{dl} \times \vec{v}) .\vec{B}$$

$$\displaystyle = ( 2\hat{k} \times ( 2 \hat{i} + 3\hat{j} + \hat{k} )).( \hat{i} + 2\hat{j})$$

$$\displaystyle = ( -6\hat{i} + 4 \hat{j} ).( \hat{i} + 2\hat{j})$$

$$\displaystyle =-6 +8 =2\: V$$
Question 2
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A magnet is made to oscillate with a particular frequency passing through a coil as shown in figure. The time variation of the magnitude of emf generated across the coil during one cycle is

A

B

C

D
Question 3
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On conducting U tube can slide inside the other as shown in figure 25.28 maintaining electrical contacts between the tubes. The magnetic field $$B$$ is perpendicular to the plane of the figure. If each tube moves towards the other at a constant speed $$v$$ then the induced emf in terms of $$B$$, $$l$$ and $$v$$ where $$l$$ is the width of each tube, will be

A
$$Blv$$

B
$$-Blv$$

C
$$2Blv$$

D
zero.

Solution

The total emf induced across each of the U tube is:$$\int_0^{\pi}BvR\sin\theta d\theta$$$$=2BvR=Bv(2R)=Bvl$$ (since $$l=2R$$)
The positive terminal of induced emf of u tube on left is at the top while that of right is at bottom.
Hence, the net emf induced is $$2Bvl$$.
Question 4
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A satellite orbiting the Earth at $$400$$ $$km$$ above the surface of the Earth has a $$2$$ $$m$$ long antenna oriented perpendicular to the Earth's surface. At the equator the Earth's magnetic field is $$8 \times 10^{-5} T$$ and is horizontal. Assuming the orbit to the circular, find emf induced across the ends of the antenna.(Given radius of earth $$R_e=6400\ Km$$)

A
$$1.3 V$$

B
$$1.2 V$$

C
$$1.0 V$$

D
$$0.12 V$$

Solution

We know
$$\cfrac{mv_0^2}{r}=\cfrac {GMm}{r^2}$$
Velocity of satellite $$v_0 =R \sqrt {\cfrac {g}{(R+h)}}=7.68\times 10^3 m/s$$
Induced $$emf = Blv_0 = 8 \times 10^{-5} \times 2 \times 7.68 \times 10^3 = 1.2 V$$
Question 5
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A long metal bar of $$30cm$$ length is aligned along a north south line and moves eastward at a speed of $$10ms^{-1}$$. A uniform magnetic field of $$4.0 T$$ points vertically downwards. If the south end of the bar has a potential of $$0 V$$, the induced potential at the north end of the bar is:

A
$$+ 12 V$$

B
$$- 12 V$$

C
$$0 V$$

D
$$cannot\ be\ determined\ since\ there\ is\ not\ closed\ circuit$$

Solution

induced emf $$= Blv =12 V$$. It is induced in the northward direction by right hand rule(emf=$$ \vec{V} \times \vec{B} $$ )
therefore if south end of pole has potential of $$0 V$$, north end will have a potential of $$12 V$$
Question 6
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The induction coil works on the principle of

A
self induction

B
mutual induction

C
amperes rule

D
fleming's right hand rule

Solution

Induction coil was the first type of transformer. It works on the principle of mutual induction.
Question 7
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The network shown in the figure is part of a complete circuit. If at a certain instant, the current $$I$$ is $$5A$$ and it is decreasing at a rate of $$10^{3} As^{-1}$$ then $$V_{B}-V_{A}$$ equals :

A
$$20 V$$

B
$$15 V$$

C
$$10 V$$

D
$$5 V$$

Solution

We will write voltage drop equation as we go from A to B.
$${ V }_{ A }-iR+V+L\dfrac { di }{ dt } ={ V }_{ B }$$
$${ V }_{ A }-5\times 1+15-5\times { 10 }^{ -3 }\times (-{ 10 }^{ 3 })={ V }_{ B }$$
$$V_B-V_A=15$$
Question 8
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Directions For Questions

Consider a toroidal solenoid with rectangular cross-section as shown in Fig. While finding the expression for magnetic field in toroid, we considered that magnetic field remains same in the whole cross-section. In actual practice we know that it varies with distance. Use Ampere's law and consider a solenoid (toroidal) having $$N$$ turns uniformly spaced with air inside. The inner radius is $$a$$ and outer radius is $$b$$. Its height is $$h$$.

...view full instructions

The self inductance of the toroidal solenoid described in the passage is

A
$$\dfrac {\mu_0N^2h}{2\pi}$$

B
$$\dfrac {\mu_0N^2h}{2\pi}log_e\dfrac {b}{a}$$

C
$$\dfrac {\mu_0Nh}{2\pi}log_e\dfrac {b}{a}$$

D
$$\dfrac {\mu_0N^2h}{2\pi}log_e\dfrac {a}{b}$$

Solution

Consider a small section of height $$h$$ and width $$dr$$ of the total toroidal cross-section.
Using Ampere's circuital law, $$\int B\cdot dl=N\mu_0i$$
Or $$B\times 2\pi r = N\mu_0 i$$
$$\implies \ B=\dfrac {\mu_0iN}{2\pi r}$$
Magnetic flux through this small section, $$d\phi=BdA$$
where $$dA = hdr$$
So, $$\phi_{per\: turn}$$ $$=\int d\phi=B\int dA$$
$$\phi_{per\: turn}$$ $$=\dfrac {\mu_0iNh}{2\pi}\int_a^b\dfrac {dr}{r}=\dfrac {\mu_0iNh}{2\pi}\ln \dfrac {b}{a}$$
Total flux $$\phi_{total}=\text {Number of turns}\times \phi_{per\:turn}$$
$$\phi_{total}$$ $$=\dfrac {\mu_0N^2hI}{2\pi}\ln \dfrac {b}{a}$$
Using $$\phi = Li$$
We get $$L=\dfrac {\mu_0N^2h}{2\pi}\ln\dfrac {b}{a}$$
Question 9
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Magnetic flux during time interval $$\tau$$ varies through a stationary loop of resistance $$R$$, as $$\phi_B = at (\tau - t)$$. Find the amount of heat generated during that time. Neglect the inductance of the loop.

A
$$\displaystyle \dfrac{a^2 \tau^3}{R}$$

B
$$\displaystyle \dfrac{a^2 \tau^2}{2R}$$

C
$$\displaystyle \dfrac{a^2 \tau^3}{3R}$$

D
$$\displaystyle \dfrac{a^2 \tau^3}{4R}$$

Solution

$$i = \displaystyle \frac{d\phi}{dt } \dfrac{1}{R} = \frac{a(\tau - 2t)}{R}$$

Heat produced $$H = \displaystyle \int_0^{\tau} i^2 Rdt$$

$$H = \displaystyle \int_0^{\tau} \frac{a^2 (\tau - 2t)^2}{R} dt = \frac{a^2 \tau^3}{3R}$$
Question 10
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Two rail tracks are $$1m$$ apart and insulated from each other and insulated from ground. A milli-voltmeter is connected across the rail-tracks. When a train travelling at 180 $$km/h$$ passes through what will be the reading in milli-voltmeter? Given : horizontal component of earth's field $$\sqrt{3} \times 10^{-4} T$$ and dip at the place $$60^o$$.

A
$$1.5 mV$$

B
$$15 mV$$

C
$$\displaystyle \dfrac{15}{\sqrt{3}} mV$$

D
$$\displaystyle \dfrac{1.5}{\sqrt{3}}mV$$

Solution

Vertical component of the magnetic field will be $$B_v$$
$$= \displaystyle B_H \tan \delta = 3 \times 10^{-4} T$$
Reading of mili voltmeter $$=B_v Iv= 3 \times 10^{-4} \times 1 \times 50 = 15 mV$$