Download App

Physics M.C.Q (1 Marks)

Electromagnetic Induction · MP Board (MPBSE) STD 12 Science (Madhya Pradesh - English Medium). 15 questions.

Questions

M.C.Q (1 Marks)

🎯

Test yourself on this topic

15 questions · timed · auto-graded

MCQ 11 Mark
A rod $AB$ moves with a uniform velocity $v$ in a uniform magnetic field as shown in figure.

  • A
    The rod becomes electrically charged.
  • ✓
    The end $A$ becomes positively charged.
  • C
    The end $B$ becomes positively charged.
  • D
    The rod becomes hot because of Joule heating.
Answer
Correct option: B.
The end $A$ becomes positively charged.
The end $'A\ '$ becomes, positively charged.
Because magnetic field exerts an average Force $\overrightarrow{\text{F}}_0=\text{q}\vec{\text{v}}\times\vec{\text{B} } n$ each free electron
where $q = 1. 6 \times 10^{19} C$ is the charge on the electron.
This Force is towards $AB$ and hence the free electrons will move towareds $B$.
Negative charge is accumulated at $'B\ '$ and positive charge appears at $A.$
View full question & answer→
MCQ 21 Mark
Consider the situation shown in figure. If the switch is closed and after some time it is opened again, the closed loop will show:
  • A
    An anticlockwise current-pulse.
  • B
    A clockwise current-pulse.
  • C
    An anticlockwise current-pulse and then a clockwise current-pulse.
  • ✓
    A clockwise current-pulse and then an anticlockwise current-pulse.
Answer
Correct option: D.
A clockwise current-pulse and then an anticlockwise current-pulse.


When the switch is closed than a clock wise current pulse generated (Because initially current flow the terminal to negative terminal).

Due to Mutual Induction, current is generated in the loop. If circuit is open after some time. Dut to loop an anticlock wise current pulse generated in the circuit.
View full question & answer→
MCQ 31 Mark
Consider the following statements:
An emf can be induced by moving a conductor in a magnetic field.
An emf can be induced by changing the magnetic field.
  • ✓
    Both A and B are true.
  • B
    A is true but B is false.
  • C
    B is true but A is false.
  • D
    Both A and B are false.
Answer
Correct option: A.
Both A and B are true.
Þ An emf con be induced by moving a condcutor in a magnetic field.
$\hat{\text{I}}=\text{Bvl}$
Þ An emf can be induced by charging the magnetic field.
$\in=\frac{-\text{d}\phi}{\text{dt}} \phi\rightarrow\text{flux}$
View full question & answer→
MCQ 41 Mark
A conducting loop is placed in a uniform magnetic field with its plane perpendicular to the field. An emf is induced in the loop if:
  • A
    It is translated.
  • B
    It is deformed.
  • C
    It is rotated about a diameter.
  • ✓
    All
Answer
Correct option: D.
All
$þ$ An emf is induced in the loop is it is rotated about a diameter
$þ$ An emf is induced in the loop if it is deformed.
View full question & answer→
MCQ 51 Mark
A small, conducting circular loop is placed inside a long solenoid carrying a current. The plane of the loop contains the axis of the solenoid. If the current in the solenoid is varied, the current induced in the loop is:
  • A
    Clockwise.
  • B
    Anticlockwise.
  • ✓
    Zero.
  • D
    Clockwise or anticlockwise depending on whether the resistance in increased or decreased.
Answer
Correct option: C.
Zero.
The angle between magnetic field and area vector is $90^\circ$, so the flux associated with coil is zero. Although magnetic field is changing but flux is remaining constant equal to zero, so emf induced and hence current in the loop is equal to zero.
View full question & answer→
MCQ 61 Mark
A conducting rod is moved with a constant velocity v in a magnetic field. A potential difference appears across the two ends:
  • A
    If $\vec{\text{v}}\parallel\vec{\text{l}}$
  • B
    If $\vec{\text{v}}\parallel\vec{\text{B}}$
  • C
    If $\vec{\text{l}}\parallel\vec{\text{B}}$
  • ✓
    None of these.
Answer
Correct option: D.
None of these.
​​​​​​​Explanation:
Potential difference appears across the two ends $=\text{Bvl}$
$\text{v}^\hat{}\text{B}, \ \text{v}^\hat{}\text{I}, \ \text{I}^\hat{}\text{B}$
View full question & answer→
MCQ 71 Mark
A conducting square loop of side l and resistance R moves in its plane with a uniform velocity u perpendicular to one of its sides. A uniform and constant magnetic field B exists along the perpendicular to the plane of the loop as shown in figure. The current induced in the loop is:
  • A
    $\frac{\text{Blv}}{\text{R}}$ clockwise.
  • B
    $\frac{\text{Blv}}{\text{R}}$ anticlockwise.
  • C
    $\frac{2\text{Blv}}{\text{R}}$ anticlockwise.
  • ✓
    $\text{Zero.}$
Answer
Correct option: D.
$\text{Zero.}$
Explanation:
Induced emf is AB is Bvl and Induced emf is DC is also Bvl.
Net emf in the closed circuit (loop) is zero.
So induced current in the loop is zero.
View full question & answer→
MCQ 81 Mark
Solve the previous question if the closed loop is completely enclosed in the circuit containing the switch.
  • A
    An anticlockwise current-pulse.
  • B
    A clockwise current-pulse.
  • ✓
    An anticlockwise current-pulse and then a clockwise current-pulse.
  • D
    A clockwise current-pulse and then an anticlockwise current-pulse.
Answer
Correct option: C.
An anticlockwise current-pulse and then a clockwise current-pulse.
An anticlockwise current-pulse generated and then a clock-wise current pulse.
View full question & answer→
MCQ 91 Mark
An LR circuit with a battery is connected at t = 0. Which of the following quantities is not zero just after the connection?
  • A
    Current in the circuit.
  • B
    Magnetic field energy in the inductor.
  • C
    Power delivered by the battery.
  • ✓
    Emf induced in the inductor.
Answer
Correct option: D.
Emf induced in the inductor.
Current will be zero at t = 0+
if the current is zero magnetic field will be zero at t = 0+
power delivered will be zero at t = 0+
The EMF induced will be equal to the applied voltage in the inductor to oppose the current.
View full question & answer→
MCQ 101 Mark
Figure shows a horizontal solenoid connected to a battery and a switch. A copper ring is placed on a frictionless track, the axis of the ring being along the axis of the solenoid. As the switch is closed, the ring will:
  • A
    Remain stationary.
  • B
    Move towards the solenoid.
  • ✓
    Move away from the solenoid.
  • D
    Move towards the solenoid or away from it depending on which terminal (positive or negative) of the battery is connected to the left end of the solenoid.
Answer
Correct option: C.
Move away from the solenoid.

$\text{e}=-\text{L}\frac{\text{di}}{\text{dt}}$

Current flow in the CKt is clock wise direction, due to Mutual Induction current flow in the loop anti clockwise direction. The net force applied on the loop in east direction. So we can say that the ring will move away from the solenoid.
View full question & answer→
MCQ 111 Mark
A rod of legth l rotates with a uniform angular velocity $\omega$ about its perpendicular bisector. A uniform magnetic field B exists parallel to the axis of rotation. The potential difference between the two ends of the rod is:
  • ✓
    $\text{zero}$
  • B
    $\frac{1}{2}\text{Bl}\omega^2$
  • C
    $\text{Bl}\omega ^2$
  • D
    $2\text{B}\text{l}\omega^2$
Answer
Correct option: A.
$\text{zero}$

Emf at both end is same $=\frac{1}{8}\text{Bwl}^2$

So the potential difference between the two ends of therod is zero.
View full question & answer→
MCQ 121 Mark
A rod of length l rotates with a small but uniform angular velocity $\omega$ about its perpendicular bisector. A uniform magnetic field B exists parallel to the axis of rotation. The potential difference between the centre of the rod and an end is:
  • A
    $\text{zero}$
  • ✓
    $\frac{1}{8}\omega\text{Bl}^2$
  • C
    $\frac{1}{2}\omega\text{Bl}^2$
  • D
    $\text{B}\omega\text{l}^2$
Answer
Correct option: B.
$\frac{1}{8}\omega\text{Bl}^2$


Take a small element dx at a distance of 'x' centre


$\text{Þ}\text{d}\in\int_{0}^{\frac{1}{2}}\text{B}\omega\text{x}\text{dx}=\frac{\text{B}\omega\text{x}^2}{2}\Big|_{0}^{\frac{1}{2}}$
$\in=\frac{1}{8}\omega\text{Bl}^2$
View full question & answer→
MCQ 131 Mark
Consider the situation shown in figure. The wire AB is slid on the fixed rails with a constant velocity. If the wire AB is replaced by a semicircular wire, the magnitude of the induced current will:
  • A
    Increase.
  • ✓
    Remain the same.
  • C
    Decrease.
  • D
    Increase or decrease depending on whether the semicircle bulges towards the resistance or away from it.
Answer
Correct option: B.
Remain the same.

E = Bvl

If the wire AB is replaced by a semicircular wire, the magnitude of the induced current will be same. Because it is depend on the velcoty & lenght of the wire.
View full question & answer→
MCQ 141 Mark
Two circular loops of equal radii are placed coaxially at some separation. The first is cut and a battery is inserted in between to drive a current in it. The current c~anges slightly because of the variation in resistance with temperature. During this period, the two loops:
  • ✓
    Attract each other.
  • B
    Repel each other.
  • C
    Do not exert any force on each other.
  • D
    Attract or repel each other depending on the sense of the current.
Answer
Correct option: A.
Attract each other.
Due to Mutual induction, current is generated in second loop and that causes the two loops attract each other.
View full question & answer→
MCQ 151 Mark
A bar magnet is released from rest along the axis of a very long, vertical copper tube. After some time the magnet:
  • A
    Will stop in the tube.
  • ✓
    Will move with almost content speed.
  • C
    Will move with an acceleration g.
  • D
    Will oscillate.
Answer
Correct option: B.
Will move with almost content speed.


After sometime the Magnet will move with almost contant speed.
View full question & answer→
Generate Paper FreeAll Electromagnetic Induction topics
M.C.Q (1 Marks) - Physics STD 12 Science Questions - Vidyadip