Electromagnetic Induction(11th And 12th > Physics ) Questions and answers for exam Preparation

11th And 12th > Physics

ELECTROMAGNETIC INDUCTION MCQs

Total Questions : 15 | Page 1 of 2 pages

A circular loop of radius R carrying current I lies in x-y plane with its centre at origin. The total magnetic flux through x-y plane is

Directly proportional to I
Directly proportional to R
Directly proportional to
Zero

Discuss Question

Answer: Option D. -> Zero
:
D

Circular loop behaves as a magnetic dipole whose one surface will be N-pole and another will be S-pole. Therefore magnetic lines a force emerges from N will meet at S. Hence total magnetic flux through x-y plane is zero.

Question 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 P (as seen by E) and an induced current $I_{Q 1}$ flows in Q. The switch remains closed for a long time. When S is opened, a current $I_{Q 2}$ flows in Q. Then the directions of $I_{Q 1}$ and $I_{Q 2}$ (as seen by E) are

Respectively clockwise and anticlockwise
Both clockwise
Both anticlockwise
Respectively anticlockwise and clockwise

Discuss Question

Answer: Option D. -> Respectively anticlockwise and clockwise
:
D

When switch S is closed magnetic field lines passing through Q increases in the direction from right to left. So, according to Lenz’s law induced current in Q i.e. $I_{Q 1}$ will flow in such a direction so that the magnetic field lines due to $I_{Q 2}$ passes from left to right through Q. This is possible when $I_{Q 1}$ flows in anticlockwise direction as seen by E. Opposite is the case when switch S is opened i.e. $I_{Q 2}$ will be clockwise as seen by E.

Question 3.

A coil of wire having finite inductance and resistance has a conducting ring placed coaxially within it. The coil is connected to a battery at time t = 0, so that a time-dependent current starts flowing through the coil. If $I_{1}$ is the current induced in the ring, and $B$ is the magnetic field at the axis of the coil due to $I_{2}$ , then as a function of time $(t > 0),$ the product $I_{2}$ (t) B(t)

Increases with time
Decreases with time
Does not vary with time
Passes through a maximum

Discuss Question

Answer: Option D. -> Passes through a maximum
:
D

Using $k_{1}, k_{2}$ etc, as different constants.
$I_{1} (t) = k_{1} [1 - e^{\frac{- t}{τ}}], B (t) = k_{2} I_{1} (t)$
$I_{2} (t) = k_{3} \frac{d B (t)}{d t} = k_{4} e^{\frac{- t}{τ}}$
$∴ l_{2} (t) B (t) = k_{5} [1 - e^{\frac{- t}{τ}}] [e^{\frac{- t}{τ}}]$
This quantity is zero for t=0 and $t = \infty$ and positive for other value of t. It must , therefore, pass through a maximum.

Question 4.

The resistance in the following circuit is increased at a particular instant. At this instant the value of resistance is 10W. The current in the circuit will be now

i = 0.5 A
i > 0.5 A
i < 0.5 A
i = 0

Discuss Question

Answer: Option B. -> i > 0.5 A
:
B

If resistance is constant $(10 Ω)$ then steady current in the circuit $i = \frac{5}{10} = 0.5$ A . But resistance is increasing it means current through the circuit start decreasing. Hence inductance comes in picture which induces a current in the circuit in the same direction of main current. So i $>$ 0.5 A.

Question 5.

A short-circuited coil is placed in a time-varying magnetic field. Electrical power is dissipated due to the current induced in the coil. If the number of turns were to be quadrupled and the wire radius halved, the electrical power dissipated would be

Halved
The same
Doubled
Quadrupled

Discuss Question

Answer: Option B. -> The same
:
B

Power $P = \frac{e^{2}}{R};$ hence $e = - (\frac{d ϕ}{d t})$ where $ϕ$ =NBA
$∴ e = - N A (\frac{d b}{d t})$ Also $R \propto \frac{1}{r^{2}}$
Where R = resistance, r = radius, l =Length
$∴ P \propto \frac{N^{2} r^{2}}{l} \Rightarrow \frac{P_{1}}{P_{2}} = 1$

Question 6.

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 is equal to

$\frac{B_{o}^{2} π r^{2}}{R}$
$\frac{B_{o} 10 r^{3}}{R}$
$\frac{B_{o}^{2} π^{2} r^{4} R}{5}$
$\frac{B_{o}^{2} π^{2} r^{4}}{R}$

Discuss Question

Answer: Option D. ->

\frac{B_{o}^{2} π^{2} r^{4}}{R}

:
D

$P = \frac{e^{2}}{R}; e = - \frac{d}{d t} (B A) = - A \frac{d}{d t} (B_{o} e^{- t}) = A B_{o} e^{- t}$
$\Rightarrow P = \frac{1}{R} (A B_{o} e^{- t})^{2} = \frac{A^{2} B_{o}^{2} e^{- 2 t}}{R}$
At the time of starting t = 0 so $P = \frac{A^{2} B_{o}^{2}}{R}$
$\Rightarrow P = \frac{(π r^{2})^{2} B_{o}^{2}}{R} = \frac{B_{o}^{2} π^{2} r^{4}}{R}$

Question 7.

A conducting ring is placed around the core of an electromagnet as shown in fig. When key K is pressed, the ring

Remain stationary
Is attracted towards the electromagnet
Jumps out of the core
None of the above

Discuss Question

Answer: Option C. -> Jumps out of the core
:
C

When key k is pressed, current through the electromagnet start increasing i.e. flux linked with ring increases which produces repulsion effect.

Question 8.

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

Discuss Question

Answer: Option C. -> 2N
:
C

Motional emf $e = B v l \Rightarrow e = 2 \times 2 \times 1 = 4$ V
This acts as a cell of emf E=4V and internal resistance r=2 $Ω$ .
This simple circuit can be drwan as follows

Current through the connector $i = \frac{4}{2 + 2} =$ 1 A
$∴$ Magnetic force on connector $F_{m} =$ =Bil= $2 \times 1 \times 1 = 2$ N (Towards left)

Question 9.

If in a coil rate of change of area is $\frac{5 m e t r e^{2}}{m i l l i s e c o n d}$ and current become 1 amp form 2 amp in $2 \times 10^{- 3}$ sec. If magnetic field is 1 Tesla then self inductance of the coil is

2 H
5 H
20 H
10 H

Discuss Question

Answer: Option D. -> 10 H
:
D

$e = B . \frac{d A}{d t} = L \frac{d i}{d t} \Rightarrow 1 \times \frac{5}{10^{- 3}} = L \times \frac{(2 - 1)}{2 \times 10^{- 3}} \Rightarrow L = 10 H$

Question 10.

A pair of parallel conducting rails lie at right angle to a uniform magnetic field of 2.0 T as shown in the fig. Two resistors 10 $Ω$ and 5 $Ω$ are to slide without friction along the rail. The distance between the conducting rails is 0.1 m. Then

Induced current = $\frac{1}{150} A$ directed clockwise if 10 $Ω$ resistor is pulled to the right with speed $0.5 m s^{- 1}$ and 5 $Ω$ resistor is held fixed
Induced current = $\frac{1}{300} A$ directed anti-clockwise if 10 $Ω$ resistor is pulled to the right with speed $0.5 m s^{- 1}$ and 5 $Ω$ resistor is held fixed
Induced current = $\frac{1}{300} A$ directed clockwise if 5 $Ω$ resistor is pulled to the left at speed $0.5 m s^{- 1}$ and 10 $Ω$ resistor is held at rest
Induced current = $\frac{1}{150} A$ directed anti-clockwise if 5 $Ω$ resistor is pulled to the left with speed $0.5 m s^{- 1}$ and 10 $Ω$ resistor is held at rest

Discuss Question

Answer: Option D. -> Induced current =

\frac{1}{150} A

directed anti-clockwise if 5

Ω

resistor is pulled to the left with speed

0.5 m s^{- 1}

and 10

Ω

resistor is held at rest
:
D

When 5 $Ω$ resistor is pulled left at 0.5 m/sec induced emf., in the said resistor
$= e = v B l = 0.5 \times 2 \times 0.1 = 0.1$ V
Resistor $10 Ω$ is at rest so induced emf in it (e=vBl) be zero.
Now net emf., in the circuit=0.1V and Equivalent resistance of the circuit
R= $15 Ω$

Hence current $i = \frac{0.1}{15}$ amp= $\frac{1}{150}$ amp
And its direction will be anti-clockwise (according to Lenz's law)