Q 1 :

The magnetic moment of a bar magnet is 0.5 Am2. It is suspended in a uniform magnetic field of 8×10-2 T. The work done in rotating it from its most stable to most unstable position is                            [2024]

  • 16×10-2J

     

  • zero

     

  • 8×10-2J

     

  • 4×10-2J

     

(3)     

         At stable equilibrium,U=-MBcos0°=-MB

          At unstable equilibrium,U=-MBcos180°=+MB

          ∆U=2MB=2×0.5×8×10-2

         Work done,W=ΔU=2MB

         W=8×10-2J

 



Q 2 :

A current of 200μA deflects the coil of a moving coil galvanometer through 60°. The current to cause deflection through π/10 radian is          [2024]

  • 30μA

     

  • 120μA

     

  • 60μA

     

  • 180μA

     

(3)   

          i∝θ(angle of deflection)

          ∴i2i1=θ2θ1⇒i2200μA=π/10π/3=310

          ⇒i2=60μA

 



Q 3 :

A coil having 100 turns, area of 5×10-3 m2, carrying current of 1 mA is placed in a uniform magnetic field of 0.20 T in such a way that plane of coil is perpendicular to the magnetic field. The work done in turning the coil through 90° is ____ μJ.                        [2024]



(100)     W=ΔU=Uf-Ui

             W=|MBcos90°-MBcos0°|=MB

             W=(NIA)B=(100×5×10-3×1×10-3)×0.2J

             =1×10-4J=100μJ

 



Q 4 :

The magnetic potential due to a magnetic dipole at a point on its axis situated at a distance of 20 cm from its center is 1.5×10-5 Tm. The magnetic moment of the dipole is _____. (Given: μ0/4π=10-7 TmA-1)                                 [2024]



(6)      The magnetic potential due to a magnetic dipole,

           V=μ04πMr2

           ⇒1.5×10-5=10-7×M(20×10-2)2

            ⇒M=1.5×10-5×20×20×10-410-7

           M=1.5×4=6



Q 5 :

A moving coil galvanometer has 100 turns and each turn has an area of 2.0 cm². The magnetic field produced by the magnet is 0.01 T and the deflection in the coil is 0.05 radian when a current of 10 mA is passed through it. The torsional constant of the suspension wire is x×10-5 N-m/rad. The value of x is _____.                                 [2024]



(4)      Given,B=0.01T,i=10mA,N=100

           A=2×10-4m2:K=?

           τ=Kθ                                           ⋯(i)

           τ=BiNAsinϕ                            ⋯(ii)

          From (i) and (ii),Kθ=BiNAsin90°

          K=BiNAθ

         K=(0.01)×(10×10-3A)×(100)×(2×10-4m2)(0.05)

         K=2×10-45=205×10-5=4×10-5N-m/rad

         x=4

 



Q 6 :

A uniform magnetic field of 2×10-3T acts along positive Y-direction. A rectangular loop of sides 20 cm and 10 cm with current of 5A is in Y-Z plane. The current is in anticlockwise sense with reference to negative X axis. Magnitude and direction of the torque is                     [2024]

  • 2×10-4 N-m along positive Z-direction

     

  • 2×10-4 N-m along negative Z-direction

     

  • 2×10-4 N-m along positive X-direction

     

  • 2×10-4 N-m along positive Y-direction

     

(2)

M→=IA→

=5×(0.2)×(0.1)(-i^)=0.1(-i^)

τ→=M→×B→=0.1(-i^)×(2×10-3)(j^)

=2×10-4(-k^) N-m



Q 7 :

A circular coil having 200 turns, 2.5×10-4m2 area and carrying 100 μA current is placed in a uniform magnetic field of 1 T. Initially, the magnetic dipole moment (M→) was directed along  B→. Amount of work, required to rotate the coil through 90° from its initial orientation such that M→ becomes perpendicular to B→, is ____ μJ.           [2024]



(5)

Work done by external magnetic field ΔU=Uf-Ui

Initially, Ui=-MBcosθ=-MBcos0°=-MB

Finally, Uf=-MBcos90°=0

The magnetic moment of coil, M=NiA

M=(100×10-6)·(200)·(2.5×10-4)

Work, W=ΔU=MB=(5×10-6)(1)=5μJ



Q 8 :

A solenoid having area A and length 'l' is filled with a material having relative permeability 2. The magnetic energy stored in the solenoid is::          [2025]

  • B2Alμ0

     

  • B2Al2μ0

     

  • B2Al

     

  • B2Al4μ0

     

(4)

uV=B22μrμ0 ⇒ u=B24μ0V=B24μ0Al



Q 9 :

A magnetic dipole experiences a torque of 803 Nm when placed in uniform magnetic field in such a way that dipole moment makes angle of 60° with magnetic field. The potential energy of the dipole is:         [2025]

  • 80 J

     

  • –403 J

     

  • –60 J

     

  • –80 J

     

(4)

|τ→|=|M→×B→|=MB sin 60°

803=32MB ⇒ MB=160

U=–M→·B→=–MB cos 60°

U=–160×12=–80 J



Q 10 :

For a moving coil galvanometer, the deflection in the coil is 0.05 rad when a current of 10 mA is passed through it. If the torsional constant of suspension wire is 4.0×10-5 Nm rad-1, the magnetic field is 0.01 T and the number of turns in the coil is 200, the area of each turn (in cm2) is          [2023]

  • 2.0

     

  • 1.0

     

  • 1.5

     

  • 0.5

     

(2)

τ=Kθ

NiAB=Kθ

A=KθNiB=4×10-5×0.05200×10×10-3×0.01

On solving, A=10-4 m2=1 cm2



Q 11 :

A square loop of area 25 cm2 has a resistance of 10 Ω. The loop is placed in uniform magnetic field of magnitude 40.0 T. The plane of loop is perpendicular to the magnetic field. The work done in pulling the loop out of the magnetic field slowly and uniformly in 1.0 sec, will be              [2023]

  • 2.5×10-3J

     

  • 1.0×10-3J

     

  • 1.0×10-4J

     

  • 5×10-3J

     

(2)

l=50 cm

t=1 s

∴  V=0.051=0.05 m/s

         i=40×0.05×0.0510=0.01 A

∴  F=BI∆Cℓ=40×0.01×0.05

         F=0.02 N

∴  W=F×ℓ=0.02×0.05

∴  W=1×10-3 J



Q 12 :

The magnetic moments associated with two closely wound circular coils A and B of radius rA=10 cm and rB=20 cm respectively are equal if (where NA,IA and NB,IB are number of turns and current of A and B respectively).              [2023]

  • 2NAIA=NBIB

     

  • NA=2NB

     

  • NAIA=4NBIB

     

  • 4NAIA=NBIB

     

(3)

M=NIA

MA=MB

∴  NAIAAA=NBIBAB

∴  NAIAπ(0.1)2=NBIBπ(0.2)2

∴  NAIA=4NBIB



Q 13 :

Given below are two statements:

Statement I: If the number of turns in the coil of a moving coil galvanometer is doubled, then the current sensitivity becomes double.

Statement II: Increasing current sensitivity of a moving coil galvanometer by only increasing the number of turns in the coil will also increase its voltage sensitivity in the same ratio.

In the light of the above statements, choose the correct answer from the options given below:                     [2023]

  • Statement I is false but Statement II is true

     

  • Both Statement I and Statement II are true

     

  • Both Statement I and Statement II are false

     

  • Statement I is true but Statement II is false

     

(4)

For a moving coil galvanometer

BiNA=kθ

θ=(BNAk)i;  current sensitivity=BNAk

So, if N is doubled then current sensitivity is doubled.

Voltage sensitivity

BVRNA=kθ

V=BNARkθ as N is doubled R is also doubled.

So, no change in voltage sensitivity.

Hence, option (4) is right.



Q 14 :

The current sensitivity of moving coil galvanometer is increased by 25%. This increase is achieved only by changing the number of turns of coils and area of cross section of the wire while keeping the resistance of galvanometer coil constant. The percentage change in the voltage sensitivity will be:              [2023]

  • +25%

     

  • −50%

     

  • Zero

     

  • −25%

     

(1)

Is=NBAC  &  Vs=NBACG

⇒  Vs=IsG

If G (galvanometer resistance) is constant, then Vs∝Is

So percentage change in Vs is also 25%.



Q 15 :

A bar magnet of dipole moment 104JT-1 is free to rotate in a horizontal plane. A horizontal magnetic field 4×10-5T exists in the space. Find the work done in rotating the magnet slowly from a direction parallel to the field to a direction 60° from the field.

  • 0.1 J

     

  • 0.2 J

     

  • 0.4 J

     

  • 0.5 J

     

(2)

W=MB(cos0°-cos60°)=MB2



Q 16 :

An insulated wire is wound so that it forms a flat coil with N = 200 turns. The radius of the innermost turn is r1=3 cm, and of the outermost turn r2=6 cm. If 20 mA current flow in it then the magnetic moment will be α×10-2 A·m2. The value of α is ________.                  [2026]

  • 4.4

     

  • 2.64

     

  • 3.25

     

  • 1.2

     

(2)

n=Nr2-r1=2003×10-2turnm

dM=ndr Iπr2

M=∫dM=nIπ∫r1r2r2 dr

M=200×20×10-3×3.14×[216-27]×10-63×10-2×3

M=2.64×10-2 A.m2



Q 17 :

A circular coil of radius 2 cm and 125 turns carries a current of 1 A. The coil is placed in a uniform magnetic field of magnitude 0.4 T. The axis of the coil makes an angle of 30° with the direction of the magnetic field. The torque acting on the coil is α×10-4 Nm. The value of α is ________.  (π=3.14)              [2026]



(314)

τ=|M→×B→|=|NIA→×B→|

τ=125×1×π(2100)2×0.4×sin30°

   =100π×10-4 N m

So α=100π=314