MCQs Of Moving Charges And Magnetism, Past Year Questions JEE Main, Class 12, Physics


52 Questions MCQ Test Class 12 Physics 35 Years JEE Mains &Advance Past year Paper | MCQs Of Moving Charges And Magnetism, Past Year Questions JEE Main, Class 12, Physics


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This mock test of MCQs Of Moving Charges And Magnetism, Past Year Questions JEE Main, Class 12, Physics for JEE helps you for every JEE entrance exam. This contains 52 Multiple Choice Questions for JEE MCQs Of Moving Charges And Magnetism, Past Year Questions JEE Main, Class 12, Physics (mcq) to study with solutions a complete question bank. The solved questions answers in this MCQs Of Moving Charges And Magnetism, Past Year Questions JEE Main, Class 12, Physics quiz give you a good mix of easy questions and tough questions. JEE students definitely take this MCQs Of Moving Charges And Magnetism, Past Year Questions JEE Main, Class 12, Physics exercise for a better result in the exam. You can find other MCQs Of Moving Charges And Magnetism, Past Year Questions JEE Main, Class 12, Physics extra questions, long questions & short questions for JEE on EduRev as well by searching above.
*Multiple options can be correct
QUESTION: 1

Advanced countries are making use of powerful electromagnets to move trains at very high speed. These trains are called maglev trains (abbreviated from magnetic levitation). These trains float on a guideway and  do not run on steel rail tracks.Instead of using a engine based on fossil fuels, they make use of magnetic field forces. The magnetized coils are arranged in the guide way which repels the strong magnets placed in the train's under carriage. This helps train move over the guideway , a technic called electro-dynamic suspension. When current passes in the coils of guideway , a typical magnetic field is set up between the undercarriage of train and guideway which pushes and pull the train along the guideway depending on the requirement.The lack of friction and its aerodynamic style allows the train to more at very high speed.

Q.1. The levitation of the train is due to

Solution:

The magnetised coils running along the track repel large magnets on the train's under carriage.

*Multiple options can be correct
QUESTION: 2

Advanced countries are making use of powerful electromagnets to move trains at very high speed. These trains are called maglev trains (abbreviated from magnetic levitation). These trains float on a guideway and  do not run on steel rail tracks.Instead of using a engine based on fossil fuels, they make use of magnetic field forces. The magnetized coils are arranged in the guide way which repels the strong magnets placed in the train's under carriage. This helps train move over the guideway , a technic called electro-dynamic suspension. When current passes in the coils of guideway , a typical magnetic field is set up between the undercarriage of train and guideway which pushes and pull the train along the guideway depending on the requirement.The lack of friction and its aerodynamic style allows the train to more at very high speed.

Q.2. The disadvantage of maglev trains is that

Solution:

Initial cost will be more.

*Multiple options can be correct
QUESTION: 3

Advanced countries are making use of powerful electromagnets to move trains at very high speed. These trains are called maglev trains (abbreviated from magnetic levitation). These trains float on a guideway and  do not run on steel rail tracks.Instead of using a engine based on fossil fuels, they make use of magnetic field forces. The magnetized coils are arranged in the guide way which repels the strong magnets placed in the train's under carriage. This helps train move over the guideway , a technic called electro-dynamic suspension. When current passes in the coils of guideway , a typical magnetic field is set up between the undercarriage of train and guideway which pushes and pull the train along the guideway depending on the requirement.The lack of friction and its aerodynamic style allows the train to more at very high speed.

Q.3. The force which makes maglev move

Solution:

The magnetic force will pull the vehicle.

*Multiple options can be correct
QUESTION: 4

The figure shows a circular loop of radius a with two long parallel wires (numbered 1 and 2) all in the plane of the paper. The distance of each wire from the centre of the loop is d. The loop and the wire are carrying the same current I.The current in the loop is in the counterclockwise direction if seen from above.

(q) The magnetic fields (B) at P due to the currents in the wires are in opposite directions.

(r) There is no magnetic field at P.

(s) The wires repel each other.

Q.4.When d ≈ a but wires are not touching the loop, it is found that the net magnetic field on the axis of the loop is zero at a height h above the loop. In that case 

Solution:

Magnetic field due to current carrying loop = Magnetic field due to straight wires

B = B1cos θ + B2cos θ = 2 B1cos θ

The current is from P to Q in wire 1 and from R to S in wire 2.

*Multiple options can be correct
QUESTION: 5

The figure shows a circular loop of radius a with two long parallel wires (numbered 1 and 2) all in the plane of the paper. The distance of each wire from the centre of the loop is d. The loop and the wire are carrying the same current I.The current in the loop is in the counterclockwise direction if seen from above.

(q) The magnetic fields (B) at P due to the currents in the wires are in opposite directions.

(r) There is no magnetic field at P.

(s) The wires repel each other.

Q.5. Consider d >> a, and the loop is rotated about its diameter parallel to the wires by 30° from the position shown in the figure. If the currents in the wires are in the opposite directions, the torque on the loop at its new position will be (assume that the net field due to the wires is constant over the loop).

Solution:

We know that torque

 

*Multiple options can be correct
QUESTION: 6

In a thin rectangular metallic strip a constant current I flows along the positive x-direction, as shown in the figure. The length, width and thickness of the strip are l, w and d, respectively.A uniform magnetic field is applied on the strip along the positive y-direction. Due to this, the charge carriers experience a net deflection along the z-direction. This results in accumulation of charge carriers on the surface PQRS and appearance of equal and opposite charges on the face opposite to PQRS. A potential difference along the z-direction is thus developed. Charge accumulation continues until the magnetic force is balanced by the electric force. The current is assumed to be uniformly distributed on the cross-section of the strip and carried by electrons.

Q.6. Consider two different metallic strips (1 and 2) of the same material. Their lengths are the same, widths are w1 and w2 and thicknesses are d1 and d2 respectively. Two points K and M are symmetrically located on the opposite faces parallel to the x-y plane (see figure). V1 and V2 are the potential differences between K and M in strips 1 and 2, respectively. Then, for a given current I flowing through them in a given magnetic field strength B, the correct statement(s) is(are)

Solution:

When megnetic force balances electric force FB = FE
q vd B = q E

   

when d1 = 2d2, V2 = 2V1

and when d1 = d2, V2 = V1 (a), (d) are correct options

 

*Multiple options can be correct
QUESTION: 7

In a thin rectangular metallic strip a constant current I flows along the positive x-direction, as shown in the figure. The length, width and thickness of the strip are l, w and d, respectively.A uniform magnetic field  is applied on the strip along the positive y-direction. Due to this, the charge carriers experience a net deflection along the z-direction. This results in accumulation of charge carriers on the surface PQRS and appearance of equal and opposite charges on the face opposite to PQRS. A potential difference along the z-direction is thus developed. Charge accumulation continues until the magnetic force is balanced by the electric force. The current is assumed to be uniformly distributed on the cross-section of the strip and carried by electrons.

Q.7. Consider two different metallic strips (1 and 2) of same dimensions (length l, width w and thickness d) with carrier densities n1 and n2, respectively. Strip 1 is placed in magnetic field B1 and strip 2 is placed in magnetic field B2, both along positive y-directions. Then V1 and V2 are the potential differences developed between K and M in strips 1 and 2, respectively. Assuming that the current I is the same for both the strips, the correct option(s) is(are)

Solution:

Here

A and C are the correct options.

*Multiple options can be correct
QUESTION: 8

Statement-1 : The sensitivity of a moving coil galvanometer is increased by placing a suitable magnetic material as a core inside the coil. and Statement-2 : Soft iron has a high magnetic permeability and cannot be easily magnetized or demagnetized. 

Solution:

Statemen t-1 is tr ue. Sensitivity =. If  B increases, θ/I
increases. Statement-2 is wrong because soft  iron can be easily magnetised and de magnetized.

 

*Multiple options can be correct
QUESTION: 9

If in a circular coil A of radius R, current I is flowing and in another coil B of radius 2R a current 2I is flowing, then the ratio of the magnetic fields BA and BB, produced by them will be

Solution:

KEY CONCEPT : We know that the magnetic field produced by a current carrying circular coil of radius r at its centre is 

*Multiple options can be correct
QUESTION: 10

If an electron and a proton having same momenta enter perpendicular to a magnetic field, then 

Solution:

KEY CONCEPT : When a charged particle enters perpendicular to a magnetic field, then it moves in a circular path of radius.

where q = Charge of the particle

p = Momentum of the particle

B = Magnetic field

Here p, q and B are constant for electron and proton, therefore the radius will be same.

*Multiple options can be correct
QUESTION: 11

Wires 1 and 2 carrying currents i1 and i2 respectively are inclined at an angle θ to each other. What is the force on a small element dl of wire 2 at a distance of r from wire 1 (as shown in figure) due to the magnetic field of wire 1? 

Solution:

Magnetic field due to current in wire 1 at point P distant r from the wire is

 (directed perpendicular to the plane  of paper, inwards)

The force exerted due to this magnetic field on current element i2 dl is

dF = i2 dl B sin 90°

*Multiple options can be correct
QUESTION: 12

The time period of a charged particle undergoing a circular motion in a uniform magnetic field is independent of its

Solution:

KEY CONCEPT : The time period of a charged particle (m, q) moving in a magnetic field (B) is T= 2πm/qB
The time period does not depend on the speed of the particle.

*Multiple options can be correct
QUESTION: 13

A particle of mass M and charge Q moving with velocity describe a circular path of radius R when subjected to a uniform transverse magnetic field of  induction B. The work done by the field when the  particle completes one full circle is

Solution:

The workdone,  dW = Fds cosθ The angle between force and displacement is 90° .
Therefore work done is zero.

*Multiple options can be correct
QUESTION: 14

A particle of charge - 16 *10 -18 coulomb moving with velocity 10ms-1 along  the x-axis enters a region where a magnetic field of induction B is along  the y-axis, and an electric field of magnitude 10 4 V / m is along the negative z-axis. If the charged particle continues moving along the x-axis, the magnitude of  B is

Solution:

The situation is shown in the figure.
FE= Force due to electric field FB = Force due to magnetic field It is given that the charged particle remains moving along X-axis (i.e. undeviated). Therefore  FB = FE

*Multiple options can be correct
QUESTION: 15

A thin rectangular magnet suspended freely has a  period of oscillation equal to T. Now it is broken into two equal halves (each having half of the original   length) and one piece is made to oscillate freely in the same field. If its period of oscillation is T ' , the ratio T '/T is

Solution:

KEY CONCEPT : The time period of a rectangular magnet oscillating in earth’s magnetic field is given by

where I = Moment of inertia of the rectangular magnetm

μ= Magnetic moment

BH = Horizontal component of the earth’s  magnetic field

Case 2 : Magnet is cut into two identical pieces such that each piece has half the original length. Then

*Multiple options can be correct
QUESTION: 16

A magnetic needle lying parallel to a magnetic field requiers W units of work to turn it through 600 . The torque needed to maintain the needle in this position will be

Solution:

W = MB(cos θ1 - cosθ2 )

= MB(cos 0° - cos 60°) 

 

*Multiple options can be correct
QUESTION: 17

The magnetic lines of force inside a bar magnet       

Solution:

As shown in the figure, the magnetic lines of force are directed from south to north inside a bar magnet.

*Multiple options can be correct
QUESTION: 18

Curie temperature is the temperature above which

Solution:

KEY CONCEPT : The temperature above which a ferromagnetic substance becomes paramagnetic is called Curie’s temperature.

*Multiple options can be correct
QUESTION: 19

A current i ampere flows along an infinitely long straight thin walled tube, then the magnetic induction at any point inside the tube is

Solution:

Using Ampere’s law at a distance r from axis, B is same from symmetry.

Here i is zero, for r < R, whereas R is the radius

∴B = 0

*Multiple options can be correct
QUESTION: 20

A long wire carries a steady current. It is bent into a circle of one turn and the magnetic field at the centre of the coil is B. It is then bent into a circular loop of n turns. The magnetic field at the centre of the coil will be

Solution:

KEY CONCEPT : Magentic field at the centre of a circular coil of radius R carrying current i is 

Given : n ´ (2pr ') =2pR

⇒ nr '=R ...(1)

from (1) and (2), 

*Multiple options can be correct
QUESTION: 21

The magnetic field due to a current carrying circular loop of radius 3 cm at a point on the axis at a distance of 4 cm from the centre is 54 μT. What will be its value at the centre of loop?

Solution:

The magnetic field at a point on the axis of a circular loop at a distance x from centre is,

*Multiple options can be correct
QUESTION: 22

Two long conductors, separated by a distance d carry current I1 and I2 in the same direction. They exert a force F on each other. Now the current in one of them is increased to two times and its direction is reversed. The distance is also increased to 3d. The new value of the force between them is

Solution:

Force between two long conductor carrying current,

*Multiple options can be correct
QUESTION: 23

The length of a magnet is large compared to its width and breadth. The time period of its oscillation in a vibration magnetometer is 2s. The magnet is cut along its length into three equal parts and these parts are then placed on each other with their like poles together. The time period of this combination will be

Solution:

When the magnet is cut into three pieces the pole strength will remain the same and

We have, Magnetic moment (M) = Pole strength (m) × l

∴  New magnetic moment,

*Multiple options can be correct
QUESTION: 24

The materials suitable for making electromagnets should have

Solution:

NOTE : Electr o mag n et sh ould be amen abl e to magnetisation &  demagnetization.

∴ retentivity should be low & coercivity should be low

*Multiple options can be correct
QUESTION: 25

Two concentric coils each of radius equal to 2 π cm are placed at right angles to each other. 3 ampere and 4 ampere are the currents flowing in each coil respectively. The magnetic induction in Weber / m2 at the centre of the coils will be

Solution:

The magnetic field due to circular coil 1 and 2 are

*Multiple options can be correct
QUESTION: 26

A charged particle of mass m and charge q travels on a circular path of radius r that is perpendicular to a magnetic field B. The time taken by the particle to complete one revolution is

Solution:

Equating magnetic force to centripetal force,

Time to complete one revolution,

*Multiple options can be correct
QUESTION: 27

A magnetic needle is kept in a non-uniform magnetic field. It experiences 

Solution:

A magnetic needle kept in non uniform magnetic field experience a force and torque due to unequal forces acting on poles.

*Multiple options can be correct
QUESTION: 28

A uniform electric field and a uniform magnetic field are acting along the same direction in a certain region. If an electron is projected along the direction of the fields with a certain velocity then 

Solution:

Due to electric field, it experiences force and decelerates i.e. its velocity decreases.

*Multiple options can be correct
QUESTION: 29

Needles N1, N2 and N3 are made of a ferromagnetic, a paramagnetic and a diamagnetic substance respectively. A magnet when brought close to them will

Solution:

Ferromagnetic substance has magnetic domains whereas paramagnetic substances have magnetic dipoles which get attracted to a magnetic field.
Diamagnetic substances do not have magnetic dipole but in the presence of external magnetic field due to their orbital motion of electrons these substances are repelled.

*Multiple options can be correct
QUESTION: 30

In a region, steady and uniform electric and magnetic fields are present. These two fields are parallel to each other. A charged particle is released from rest in this region. The path of the particle will be a

Solution:

The charged particle will move along the lines of electric field (and magnetic field). Magnetic field will exert no force. The force by electric field will be along the lines of uniform electric field. Hence the particle will move in a straight line.

*Multiple options can be correct
QUESTION: 31

A long solenoid has 200 turns per cm and carries a current i. The magnetic field at its centre is 6.28 × 10–2 Weber/m2. Another long solenoid has 100 turns per cm and it carries a current i/3 . The value of the magnetic field at its centre is

Solution:

*Multiple options can be correct
QUESTION: 32

A long straight wire of radius a carries a steady current i. The current is uniformly distributed across its cross section. The ratio of the magnetic field at a/2 and 2a is

Solution:

Here, current is uniformly distributed across the crosssection of the wire, therefore, current enclosed in the amperean path formed at a distance 

 where I is total current

∴ Magnetic field at P1 is

Now, magnetic field at point P2,

*Multiple options can be correct
QUESTION: 33

A current I flows along the length of an infinitely long, straight, thin walled pipe. Then

Solution:

There is no current inside the pipe. Therefore

*Multiple options can be correct
QUESTION: 34

A charged particle with charge q enters a region of constant, uniform and mutually orthogonal fields  with a velocity  and comes out without any change in magnitude or direction .Then

 

Solution:

Here, are perpendicular to each other and the velocity does not change; therefore

Also,

*Multiple options can be correct
QUESTION: 35

A charged particle moves through a magnetic field perpendicular to its direction. Then

Solution:

NOTE : When a charged particle enters a magnetic field at a direction perpendicular to the direction of motion, the path of the motion is circular. In circular motion the direction of velocity changes at every point (the magnitude remains constant).
Therefore, the tangential momentum will change at every point. But kinetic energy will remain constant as it is given by

1/2 mv2 and v2 is the squar e of th e magnitude of velocity which does not change.

 

*Multiple options can be correct
QUESTION: 36

Two identical conducting wires AOB and COD are placed at right angles to each other. The wire AOB carries an electric current I1 and COD carries a current I2. The magnetic field on a point lying at a distance d from O, in a direction perpendicular to the plane of the wires AOB and COD, will be given by

Solution:

Clearly, the magnetic fields at a point P, equidistant from AOB and COD will have directions perpendicular to each other, as they are placed normal to each other.

*Multiple options can be correct
QUESTION: 37

A horizontal overhead powerline is at  height of 4m from the ground and carries a current of 100A from east to west. The magnetic field directly below it on the ground is

Solution:

The magnetic field is

= 5 × 10–6 T

According to right hand palm rule, the magnetic field is directed  towards south.

*Multiple options can be correct
QUESTION: 38

Relative permittivity and permeability of a material εr and μr , respectively. Which of the following values of these quantities are allowed for a diamagnetic material?    

Solution:

For a diamagnetic material, the value of µr is less than one. For any material, the value of ∈r is always greater than 1.

*Multiple options can be correct
QUESTION: 39

A current loop ABCD is held fixed on the plane of the paper as shown in the figure. The arcs BC (radius = b) and DA (radius = a) of the loop are joined by two straight wires AB and CD.  A steady current I is flowing in the loop. Angle made by AB and CD at the origin O is 30°. Another straight thin wire with steady current I1 flowing out of the plane of the paper is kept at the origin.

Q.31. The magnitude of the magnetic field (B) due to the loop ABCD at the origin (O) is :

Solution:

The magnetic field at O due to current in DA is

 (directed vertically upwards)

The magnetic field at O due to current in BC is

(directed vertically downwards)

The magnetic field due to current AB and CD at O is zero.
Therefore the net magnetic field is B = B1-B2 (directed vertically upwards)

 

 

*Multiple options can be correct
QUESTION: 40

A current loop ABCD is held fixed on the plane of the paper as shown in the figure. The arcs BC (radius = b) and DA (radius = a) of the loop are joined by two straight wires AB and CD.  A steady current I is flowing in the loop. Angle made by AB and CD at the origin O is 30°. Another straight thin wire with steady current I1 flowing out of the plane of the paper is kept at the origin.

Q.32.Due to the presence of the current I1 at the origin:

Solution:

The force on AD and BC due to current  I1 is zero. This is because the directions of current element  and magnetic field 

*Multiple options can be correct
QUESTION: 41

Two long parallel wires are at a distance 2d apart. They carry steady equal currents flowing out of the plane of the paper as shown. The variation of the magnetic field B along the line XX' is given by

Solution:

The magnetic field varies inversely with the distance for a long conductor. That is   According to the magnitude and direction shown graph (1) is the correct one.

*Multiple options can be correct
QUESTION: 42

A current I flows in an infinitely long wire with cross section in the form of a semi-circular ring of radius R. The magnitude of the magnetic induction along its axis is:

Solution:

Current in a small element, 

Magnetic field due to the element

The component dB cos θ, of the field is cancelled by another opposite component.
Therefore,

*Multiple options can be correct
QUESTION: 43

A charge Q is uniformly distributed over the surface of nonconducting disc of radius R. The disc rotates about an axis perpendicular to its plane and passing through its centre with an angular velocity ω. As a result of this rotation  a magnetic field of induction B is obtained at the centre of the disc. If we keep both the amount of charge placed on the disc and its angular velocity to be constant and vary the radius of the disc then the variation of the magnetic induction at the centre of the disc will be represented by the figure :

Solution:

 The magnetic field due a disc is given as

*Multiple options can be correct
QUESTION: 44

Proton, deuteron and alpha particle of same kinetic energy are moving in circular trajectories in a constant magnetic field. The radii of proton, deuteron and alpha particle are respectively rp, rd and rα. Which one of the following relation is correct?

Solution:

Thus we have,  

*Multiple options can be correct
QUESTION: 45

Two short bar magnets of length 1 cm each have magnetic moments 1.20 Am2 and 1.00 Am2 respectively. They are placed on a horizontal table parallel to each other with their N poles pointing towards the South. They have a common magnetic equator and are separated by a distance of 20.0 cm. The value of the resultand horizontal magnetic induction at the mid-point O of the line joining their centres is close to (Horizontal component of earth.s magnetic induction is 3.6× 10.5Wb/m2)

Solution:

Given : 2 M1 = 1.20 Am2 and M2 = 1.00 Am2

Bnet = B1 + B2 + BH

*Multiple options can be correct
QUESTION: 46

A conductor lies along the z-axis at  and carries a fixed current of 10.0 A in   direction (see figure).For a field  T, find the power required to move the conductor at constant speed to x = 2.0 m, y = 0 m in 5*10-3s. Assume parallel motion along the x-axis. 

Solution:

Work done in moving the conductor is,

(By exponential function)

= 9 × 10–3 × (0.33) = 2.97 × 10–3J
Power required to move the conductor is,

*Multiple options can be correct
QUESTION: 47

The coercivity of a small magnet where the ferromagnet gets demagnetized is 3*103Am-1. The current required to be passed in a solenoid of length 10 cm and number of turns 100, so that the magnet gets demagnetized when inside the solenoid, is:

Solution:

Magnetic field in solenoid B =μ0ni

(Where n = number of turns per unit length)

*Multiple options can be correct
QUESTION: 48

Two long current carrying thin wires, both with current I, are held by insulating threads of length L and are in equilibrium as shown in the figure, with threads making an angle 'θ' with the vertical. If wires have mass λ per unit length then the value of I is :

(g = gravitational acceleration)

Solution:

Let us consider 'l' length of current carrying wire.
At equilibrium

 

*Multiple options can be correct
QUESTION: 49

A rectangular loop of sides 10 cm and 5 cm carrying a current 1 of 12 A is placed in different orientations as shown in the figures below :

(a)(b) 

(c) (d) 

If there is a uniform magnetic field of 0.3 T in the positive z direction, in which orientations the loop would be in (i) stable equilibrium and (ii) unstable equilibrium ?

Solution:

For stable equilibrium 

For unstable equilibrium 

 

*Multiple options can be correct
QUESTION: 50

Two identical wires A and B, each of length 'l', carry the same current I. Wire A is bent into a circle of radius R and wire B is bent to form a square of side 'a'. If BA and BB are the values of magnetic field at the centres of the circle and square respectively, then the ratio 

Solution:

Case (b) :

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QUESTION: 51

A galvanometer having a coil resistance of 100 Ω gives a full scale deflection, when a currect of 1 mA is passed through it. The value of the resistance, which can convert this galvanometer into ammeter giving a full scale deflection for a current of 10 A, is :

Solution:

Ig G = ( I – Ig)s

∴ 10–3 × 100 = (10 – 10–3) × S

∴  S ≈ 0.01 Ω

*Multiple options can be correct
QUESTION: 52

Hysteresis loops for two magnetic materials A and B are given below :

These materials are used to make magnets for elecric generators, transformer core and electromagnet core. Then it is proper to use :

 

Solution:

Graph [A] is for material used for making permanent magnets (high coercivity)

Graph [B] is for making electromagnets and transformers.