Electromagnetic Induction Test

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Electromagnetic Induction Test - 30

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Question 1
1 / -0

A conducting square loop of side $$l$$ and resistance R moves in its plane with a uniform velocity $$v$$ perpendicular to one of its sides. A uniform and constant magnitude field B exists along the perpendicular to the plane of the loop as shown in figure. The current induced in the loop is

A
$$\displaystyle \frac{Blv}{R}$$ clockwise

B
$$\displaystyle \frac{Blv}{R}$$ anti-clockwise

C
$$\displaystyle \frac{2 Blv}{R}$$ anti-clockwise

D
zero

Solution

Flux at any point in time across the loop = $$ B l^2$$
the flux does not change with time, hence emf is not induced , hence no current flows.
Question 2
1 / -0

A tank containing a liquid has turns of wire wrapped around it, causing it to act as an inductor. The liquid content of the tank can be measured by using its inductance to determine the height of the liquid level in the tank. The inductance of the tank changes from a value of $$L_0$$ corresponding to a relative permeability of $$1$$ when the tank is empty to value $$L_f$$ corresponding to a relative permeability $$X_m$$ (relative permeability of liquid) when the tank is full. The appropriate electronic circuit can determine the inductance correct upto $$5$$ significant figures and thus the effective relative permeability of the combined air and liquid within the rectangular has height $$D$$. The height of the liquid level in the tank is $$d$$. Ignore the fringing effects. Assume tank is fitted with $$Hg X _{Hg} = 2.9 \times 10^5$$.
Express $$d$$ as a function of $$L$$, inductance corresponding to a certain liquid height $$L_0, L_f$$ and $$D$$.

A
$$\displaystyle d =\frac{(L - L_0)D}{L_f - L_0}$$

B
$$\displaystyle d =\frac{LD}{L_f - L_0}$$

C
$$\displaystyle d =\frac{(L - L_0)D}{L_f }$$

D
None of these

Solution

Inductance of empty tank=$$L_0$$
Relative permeability=1
Inductance of full tank=$$L_f$$
Relative permeability=$$X_m$$
Now,$$L_f-L_0 \quad \alpha D$$
and $$L_f-L_0 \quad\alpha \dfrac{1}{d}$$
therefore, $$(L_f-L_0) \quad \alpha \dfrac{D}{d}$$
or$$(L_f-L_0)=k\dfrac{D}{d}$$ [k is the probability constant]
Now when the tank is filled partly with liquid and partly empty,Inductance=$$L$$
$$K=L-L_0$$
$$L_f-L_0=(L-L_0)\dfrac{D}{d}$$
$$d=\dfrac{(L-L_0)D}{(L_f-L_0)}$$
Question 3
1 / -0

Two conducting rings of radii r and 2r move in opposite directions with velocities 2$$v$$ and $$v$$ respectively on a conducting surface $$S$$. There is a uniform magnetic field of magnitude $$B$$ perpendicular to the plane of the rings. The potential difference between the highest points of the two rings is

A
$$zero$$

B
$$2rvB$$

C
$$4 rvB$$

D
$$8 rvB$$

Solution

Replace the induced emf in the rings by cells.
$$\varepsilon_1 = B2r (2v) = 4 Brv.$$
$$\varepsilon_2 = B (4r) v = 4 Brv$$.
$$V_2 - V_1 = \varepsilon_2 + \varepsilon_1 = 8 Brv$$.
Question 4
1 / -0

A coil of insulating wire is connected to battery. If it is moved towards a galvanometer then its point gets deflected because

A
the coil behaves like a magnet

B
induced current is produced in the coil

C
the number of turns in the galvanometer coil remains constant

D
none of the above

Solution

Upon connecting to the battery , a current starts to flow in the insulated wire. This current induces a current in the galvanometer coil due to mutual inductance because of the flux linkages. This induced current causes a deflection in the galvanometer .
Question 5
1 / -0

A 1.2 m wide railway track is parallel to magnetic meridian. The vertical component of earth's magnetic field is 0.5 Gauss. When a train runs on the rails at a speed of 60 Km/hr, then the induced potential difference the ends of its axle will be

A
$$10^{-4}$$ V

B
$$2 \times 10^{-4}$$ V

C
$$10^{-3}$$ V

D
zero

Solution

Given : $$L = 1.2$$ m $$B = 0.5\times 10^{-4}$$ T

Velocity of train $$v = 60$$ km/hr $$ = 60\times \dfrac{5}{18}$$ m/s $$ = 16.67$$ m/s

$$\therefore$$ Emf induced $$\mathcal{E} = BvL = 0.5\times 10^{-4}\times 16.67\times 1.2 = 10^{-3}$$ V
Question 6
1 / -0

Two coils P and Q are lying parallels and very close to each other. Coil P is connected to an AC source whereas Q is connected to a sensitive galvanometer. On pressing key K

A
small variations are observed in the galvanometer for applied 50 Hz voltage

B
deflections in the galvanometer can be observed for applied voltage of 1 Hz to 2 Hz.

C
no deflection in the galvanometer will be observed

D
constant deflection will be observed in the galvanometer for 50 Hz supply voltage

Solution

Upon connecting to the AC source , a current starts to flow in the coil P. This current induces a current in the galvanometer coil Q due to mutual inductance because of the flux linkages. This induced current causes a deflection in the galvanometer.
Question 7
1 / -0

When the number of turns per unit length in a solenoid is doubled then its coefficient of self induction will become

A
half

B
double

C
four times

D
unchanged

Solution

Coefficient of Self inductance is directly proportional to square of number of turns in coil, hence on doubling the number of turns. Coefficient of Self inductance becomes four times.
Question 8
1 / -0

The number of turns in an air core solenoid of length 25 cm and radius 4 cm is 100. Its self inductance will be

A
$$5 \times 10^{-4} H$$

B
$$2.5 \times 10^{-4} H$$

C
$$5.4 \times 10^{-3} H$$

D
$$2.5 \times 10^{-3} H$$

Solution

Inducatnce = $$ \mu N^2 \dfrac{ Area}{length} $$

$$= \mu 100^2 \dfrac{ \pi 0.4^2}{0.25}= 2.5 \times 10^{-4} H $$
Question 9
1 / -0

A cycle wheel with 64 spokes is rotating with N rotations per second at right angles to horizontal component of magnetic field. The induced emf generated between its axle and rim is E. If the number of spokes is reduced to 32 then the value of induced emf will be :

A
E

B
2E

C
$$\displaystyle \frac{E}{2}$$

D
$$\displaystyle \frac{E}{4}$$

Solution

The emf induced is independent of the number of spoke and only depends on the lenght of spoke and the angular speed.
Therefore emf induced is same as before = $$E$$
Question 10
1 / -0

The coefficient of mutual induction between two coils is 4 H. If the current in the primary reduces from 5A to zero in 10$$^{-3}$$ second then the induced emf in the secondary coil will be

A
$$10^4$$ V

B
$$25 \times 10^3$$ V

C
$$2 \times 10^4 $$ V

D
$$15 \times 10^3$$ V

Solution

emf = $$ - M \dfrac{dI}{dt}$$

$$ = -4 \dfrac{0-5}{10^{-3} }= 2 \times 10^4 V $$

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Electromagnetic Induction Test - 30

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Electromagnetic Induction Test - 30

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Question 1

$$\displaystyle \frac{Blv}{R}$$ clockwise

$$\displaystyle \frac{Blv}{R}$$ anti-clockwise

$$\displaystyle \frac{2 Blv}{R}$$ anti-clockwise

zero

Solution

Question 2

$$\displaystyle d =\frac{(L - L_0)D}{L_f - L_0}$$

$$\displaystyle d =\frac{LD}{L_f - L_0}$$

$$\displaystyle d =\frac{(L - L_0)D}{L_f }$$

None of these

Solution

Question 3

$$zero$$

$$2rvB$$

$$4 rvB$$

$$8 rvB$$

Solution

Question 4

the coil behaves like a magnet

induced current is produced in the coil

the number of turns in the galvanometer coil remains constant

none of the above

Solution

Question 5

$$10^{-4}$$ V

$$2 \times 10^{-4}$$ V

$$10^{-3}$$ V

zero

Solution

Question 6

small variations are observed in the galvanometer for applied 50 Hz voltage

deflections in the galvanometer can be observed for applied voltage of 1 Hz to 2 Hz.

no deflection in the galvanometer will be observed

constant deflection will be observed in the galvanometer for 50 Hz supply voltage

Solution

Question 7

half

double

four times

unchanged

Solution

Question 8

$$5 \times 10^{-4} H$$

$$2.5 \times 10^{-4} H$$

$$5.4 \times 10^{-3} H$$

$$2.5 \times 10^{-3} H$$

Solution

Question 9

E

2E

$$\displaystyle \frac{E}{2}$$

$$\displaystyle \frac{E}{4}$$

Solution

Question 10

$$10^4$$ V

$$25 \times 10^3$$ V

$$2 \times 10^4 $$ V

$$15 \times 10^3$$ V

Solution

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