Q 1 :

The electrostatic force (F→1) and magnetic force (F→2) acting on a charge q moving with velocity v can be written             [2024]

  • F→1=qB→,F→2=q(B→×v→)

     

  • F→1=qE→,F→2=q(v→×B→)

     

  • F→1=qv→·E→,F→2=q(B→·v→)

     

  • F→1=qE→,F→2=q(B→×v→)

     

(2)     

         The electrostatic force acting on a charge, F→1=qE→

          The magnetic force acting on a charge F→2=q(v→×B→)

 



Q 2 :

A proton and a deuteron (q=+e,m=2.0u) having same kinetic energies enter a region of uniform magnetic field B→ , moving perpendicular to B→. The ratio of the radius rd of deuteron path to the radius rp of the proton path is:                      [2024]

  • 1:1

     

  • 1:2

     

  • 2:1

     

  • 1:2

     

(3)     

           As radius of circular path (r)=mvqB                               ⋯(i)

           Given (KE)p=(KE)d

            12mpvp2=12mdvd2⇒mpvp2=mdvd2

            As md=2mp⇒mpvp2=2mpvd2⇒vp=2vd

           Now from equation (i),

           ⇒rprd=mpvpeB×eB2mpvd=12=rd:rp=2:1

 



Q 3 :

A proton moving with a constant velocity passes through a region of space without any change in its velocity. If E¯ and B¯ represent the electric and magnetic fields respectively, then the region of space may have:

(A) E = 0, B = 0                                (B) E = 0, B ≠ 0

(C) E ≠ 0, B = 0                                (D) E ≠ 0, B ≠ 0

Choose the most appropriate answer from the options given below:                                   [2024]

  • (A), (B) and (C) only

     

  • (A), (C) and (D) only

     

  • (A), (B) and (D) only

     

  • (B), (C) and (D) only

     

(3)      

          Net force on particle must be zero i.e.

           qE→+qV→×B→=0

           Possible cases are

           (i)  E→ and B→=0

           (ii)  V→×B→=0,E→=0

           (iii)  qE→=-qV→×B→

            E→≠0 and B→≠0

 



Q 4 :

Two particles X and Y having equal charges are being accelerated through the same potential difference. Thereafter they enter normally in a region of uniform magnetic field and describes circular paths of radii R1 and R2 respectively. The mass ratio of X and Y is                    [2024]

  • (R1R2)

     

  • (R2R1)2

     

  • (R2R1)

     

  • (R1R2)2

     

(4)   

         R=mvqB=pqB=2m(KE)qB=2mqVqB

          R∝m⇒m∝R2

          m1m2=(R1R2)2

 



Q 5 :

An electron with kinetic energy 5 eV enters a region of uniform magnetic field of 3 μT perpendicular to its direction. An electric field E is applied perpendicular to the direction of velocity and magnetic field. The value of E, so that the electron moves along the same path, is ____ NC-1. (Given, mass of electron =9×10-31 kg, electric charge =1.6×10-19C)                         [2024]



(4)            For the given condition of moving undeflected, net force should be zero.

                qE=qVB⇒E=VB

                E=2×KEm×B=2×5×1.6×10-199×10-31×3×10-6=4 N/C

 



Q 6 :

An electron moves through a uniform magnetic field B→=B0i^+2B0J^T. At a particular instant of time, the velocity of electron is u→=3i^+5j^ m/s. If the magnetic force acting on electron is F→=5ek^ N, where e is the charge of electron, then the value of B0 is ____ T.                    [2024]



(5)        F→=q(v→×B→)

             5ek^=e(3i^+5j^)×(B0i^+2B0j^)

             5ek^=e(6B0k^-5B0k^)⇒B0=5T

 



Q 7 :

An electron is projected with uniform velocity along the axis inside a current-carrying long solenoid. Then                      [2024]

  • the electron will continue to move with uniform velocity along the axis of the solenoid

     

  • the electron will be accelerated along the axis

     

  • the electron path will be circular about the axis

     

  • the electron will experience a force at 45° to the axis and execute a helical path

     

(1)

The force on a charged particle moving in a magnetic field, F→=q(v→×B→)

Since v→∥B→ so the force on the electron due to the magnetic field is zero.

So it will move along the axis with uniform velocity.



Q 8 :

Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A) : An electron in a certain region of uniform magnetic field is moving with constant velocity in a straight line path.

Reason (R) : The magnetic field in that region is along the direction of velocity of the electron.

In the light of the above statements, choose the correct answer from the options given below:           [2025]

  • (A) is false but (R) is true

     

  • Both (A) and (R) are true and (R) is the correct explanation of (A)

     

  • Both (A) and (R) are true but (R) is NOT the correct explanation of (A)

     

  • (A) is true but (R) is false

     

(2)

F→=q(v→×B→)

F→=0 when v→∥B→

In this case, θ=0° or 180°



Q 9 :

Consider a long thin conducting wire carrying a uniform current I. A particle having mass "M" and charge "q' is released at a distance "a" from the wire with a speed v0 along the direction of current in the wire. The particle gets attracted to the wire due to magnetic force. The particle turns round when it is at distance x from the wire. The value of x is [μ0 is vacuum permeability]          [2025]

  • a[1–mv02qμ0I]

     

     

  • a2

     

  • a[1–mvαqμ0I]

     

  • ae–4πmv0qμ0I

     

(4)

Let the velocity of the particle make an angle θ with initial direction when it is at a distance x.

⇒ dxdt=–v0 sin θ          ... (i)

Also, dθdt=qm(μ0I2πx)          ... (ii)

⇒ dxdθ=–2πmv0x sin θqμ0I          ... dividing eq (i) with eq (ii)

or ∫axdxx=–2πmv0qμ0I∫0πsin θdθ

lnxa=–4πmv0qμ0I

Or x=ae–4πmv0μ0Iq



Q 10 :

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.

Assertion A : If oxygen ion (O–2) and Hydrogen ion (H+) enter normal to the magnetic field with equal momentum, then the path of O–2 ion has a smaller curvature than that of H+.

Reason R : A proton with same linear momentum as an electron will from a path of smaller radius of curvature on entering a uniform magnetic field perpendicularly.

In the light of the above statement, choose the correct answer from the options given below:          [2025]

  • A is true but R is false

     

  • Both A and R true but R is NOT the correct explanation of A

     

  • A is false but R is true

     

  • Both A and R true and R is the correct explanation of A

     

(1)

Radius in magnetic field, R=mvqB

(A) mv, B are same for O2– and H+ therefore R∝1|q| therefore (A) is correct

(B) R=mvqB; mv, q, B are same, therefore (R) is incorrect.



Q 11 :

Uniform magnetic fields of different strengths ( B1and B2) both normal to the plane of the paper exist as shown in the figure. A charged particle of mass m and charge q, at the interface at an instant, moves into the region 2 with velocity v and returns to the interface. It continues to move into region 1 and finally reaches the interface. What is the displacement of the particle during this movement along the interface?

(Consider the velocity of the particle to be normal to the magnetic field and B2>B1)          [2025]

  • mvqB1(1–B2B1)×2

     

  • mvqB1(1–B1B2)

     

  • mvqB1(1–B2B1)

     

  • mvqB1(1–B1B2)×2

     

(4)

As v→ is ⊥ to B→, so charge particle will move in circular path, whose radius is given by

R=mvqB

Starting point →A

Ending point →C

∴ Net displacement = AC

AC = CD – AD

AC=2mvqB1–2mvqB2

AC=2mvqB1[1–B1B2]



Q 12 :

A particle of charge q, mass m and kinetic energy E enters in magnetic field perpendicular to its velocity and undergoes a circular arc of radius (r). Which of the following curves represents the variation of r with E?          [2025]

  •  

  •  

  •  

  •  

(4)

12mv2=E ⇒ (mv)2=2mE

Also, r=mvqB ⇒ r=2mEqB

So, r∝(E)12

Graph should be like



Q 13 :

A proton is moving undeflected in a region of crossed electric and magnetic fields at a constant speed of 2×105 ms–1. When the electric field is switched off, the proton moves along a circular path of radius 2 cm. The magnitude of electric field is x×104 N/C. the value of x is ________. (Take the mass of the proton =1.6×10–27 kg.)         [2025]



(2)

For uniform speed : Bvq = Eq ⇒ E = Bv

r=vmBq ⇒ B=mvrq

E=(mvrq)v=mv2rq

E=1.6×10–27×4×10102×10–2×1.6×10–19=2×104 N/C

⇒ x=2



Q 14 :

A tightly wound long solenoid carries a current of 1.5 A. An electron is executing uniform circular motion inside the solenoid with a time period of 75 ns. The number of turns per metre in the solenoid is ________.

[Take mass of electron me=9×10–31 kg, charge of electron |qe|=1.6×10–19C, μ0=4π×10–7NA2, 1 ns=10–9 s]          [2025]



(250)

Since time period of a revolving charge is 2πmqB

Where B = magnetic field

Due to a solenoid B = μ0nI

⇒ 2πmqT=μ0nI

⇒ n=2πmqTμ0I

            =2π×9×10–311.6×10–19×75×10–9×4π×10–7×1.5

⇒ n=9×10–311.6×10–19×75×10–9×2×10–7×1.5

⇒ n=9×104360=250 per meter



Q 15 :

A particle of charge 1.6 μC and mass 16 μg is present in a strong magnetic field of 6.28 T. The particle is then fired perpendicular to magnetic field. The time required for the particle to return to original location for the first time is __________ ms. (π = 3.14)          [2025]



(10)

Angle between v→ of charge and B→ is 90° motion will be uniform circular motion time period is given by

T=2πmqB=2π×16×10–9 kg1.6×10–6×6.28 = 0.01 seconds



Q 16 :

An electron is allowed to move with constant velocity along the axis of current carrying straight solenoid.       [2023]

A. The electron will experience magnetic force along the axis of the solenoid.
B. The electron will not experience magnetic force.
C. The electron will continue to move along the axis of the solenoid.
D. The electron will be accelerated along the axis of the solenoid.
E. The electron will follow parabolic path inside the solenoid.

Choose the correct answer from the options given below:

  • B, C and D only

     

  • B and C only

     

  • A and D only

     

  • B and E only

     

(2)

F→=q(v→×B→) as the angle between v→ and B→ is 0°

 F→=0



Q 17 :

A charge particle moving in magnetic field B, has the components of velocity along B as well as perpendicular to B. The path of the charge particle will be               [2023]

  • straight along the direction of magnetic field B

     

  • helical path with the axis perpendicular to the direction of magnetic field B

     

  • helical path with the axis along magnetic field B

     

  • circular path

     

(3)

Due to the component of velocity (v1) along the magnetic field, no force will be experienced by the particle.

So v1 remains unchanged.

But due to the component v2 magnetic force acts towards the centre, i.e., moving it in a circular path. So the path is helical with the axis parallel to the magnetic field B.



Q 18 :

A charge particle of 2 μC accelerated by a potential difference of 100 V enters a region of uniform magnetic field of magnitude 4 mT at right angle to the direction of field. The charge particle completes a semicircle of radius 3 cm inside magnetic field. The mass of the charge particle is ________ ×10-18 kg.            [2023]



(144)

r=mvqB=2kmqB

m=r2q2B22k

m=1100×3100×2×2×4×10-3×4×10-3×10-122×(100)×10-6

=144×10-18 kg



Q 19 :

A proton with a kinetic energy of 2.0 eV moves into a region of uniform magnetic field of magnitude π2×10-3 T. The angle between the direction of magnetic field and velocity of proton is 60°. The pitch of the helical path taken by the proton is _______cm. (Take, mass of proton =1.6×10-27 kg and charge on proton =1.6×10-19 C.)                     [2023]



(40)

B=π2×10-3

K.E.=12mv2  ⇒  v=2 K.E.m

Pitch=v cos 60°×time period of one rotation

           =v cos 60°×2πmeB

           =2×2×1.6×10-191.6×10-27×cos60°×2π×1.6×10-271.6×10-19×π2×10-3

            =2×104×12×4×10-5

             =4×10-1 m=40 cm



Q 20 :

A current carrying solenoid is placed vertically and a particle of mass m with charge Q is released from rest. The particle moves along the axis of solenoid. If g is acceleration due to gravity then the acceleration (a) of the charged particle will satisfy :   [2026]

  • a > g

     

  • a = g

     

  • 0 < a < g

     

  • a = 0

     

(2)

Since the solenoid is placed vertically, the magnetic field inside the solenoid will be either along -y or +y axis.

⇒  Particle will gain velocity along -y axis.

⇒  F→B=q(v→×B→)

⇒  F→B=0

⇒  F→net=mg→

⇒  anet=g



Q 21 :

Given below are two statements: One is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A): In a static magnetic field, the velocity of a charged particle changes only in direction.

Reason (R): Moving charged particle experiences magnetic force which is perpendicular to its direction of velocity.

  • Both A and R are true and R is the correct explanation of A

     

  • Both A and R are true but R is not the correct explanation of A

     

  • A is true but R is false

     

  • A is false but R is true

     

(1)

Both A and R are true and R is the correct explanation of A



Q 22 :

A proton, a deuteron and an α particle with the same KE enter in a region of uniform magnetic field, moving at right angles to magnetic field. What is the ratio of the radii of their circular paths?

  • 1:2:1

     

  • 1:2:2

     

  • 2:1:1

     

  • 2:2:1

     

(1)

rp=2mpKeB    rd=2(2mp)KeB    ra=2(4mp)K(2e)B



Q 23 :

In a coordinate system an electric field E and a magnetic field B exist in X-direction and particle of charge q and mass m is projected with a velocity v from origin along Y-direction. A target is located at point (x0,0,0). Find the magnitude of magnetic induction B (in Tesla) so that the particle will hit the target when it is crossing the X-axis third time. [Consider that π2=10, E=1, x0=1, m=2, q=10 all quantities in SI Unit]



(6)

Motion will be helical with increasing pitch, particle should hit the target in third round

x0=12qEm(3T)2

⇒B=6T



Q 24 :

When a proton is released from rest in a room, it starts with an initial acceleration a0 towards west. When it is projected towards north with a speed v0, it moves with an initial acceleration 3a0 towards west. The electric and magnetic fields in the room are

  • ma0e east, 3ma0ev0 down

     

  • ma0e west, 2ma0ev0 up

     

  • ma0e west, 2ma0ev0 down

     

  • ma0e east, 3ma0ev0 up

     

(3)

Acceleration of charged particle      a→=qm(E→+v→×B→)

Released from rest⇒ a→=qmE→=a0 (west)    ⇒    E→=ma0e (west)

When it is projected towards north, acceleration due to magnetic force=2a0

Therefore magnetic field=2ma0ev0 (down),    

Direction of magnetic field F→m→(v→×B→)-i^→(j^×-k^), 

 The direction of magnetic force is towards west, for that the direction of magnetic field is vertically downward (-k^)

 



Q 25 :

A particle having charge 10-9 C moving in x-y plane in fields of 0.4j^ N/C and 4×10-3k^ T experiences a force of (4i^+2j^)×10-10 N. The velocity of the particle at that instant is ______ m/s.                      [2026]

  • 50i^+100j^

     

  • 100i^+50j^

     

  • -50i^+100j^

     

  • 50i^-100j^

     

(1)

F→=qE→+q(v→×B→)

(4i^+2j^)×10-10=10-9[0.4j^+(vxi^+vyj^)×(4×10-3)k^]

(4i^+2j^)×10-10=10-9(0.4j^-4×10-3vxj^+4×10-3vyi^)

On comparing: vy=100, vx=50

v→=(50i^+100j^) ms-1



Q 26 :

A particle of charge q and mass m is projected from origin with an initial velocity v→=(v02x^+v02y^). There exists a uniform magnetic field B→=B0z^ and a space varying electric field E→=E0e-λxx^ within the region 0≤x≤L. After travelling a distance such that x-coordinate has changed from x=0 to x=L, the change in the kinetic energy is ________.                       [2026]

  • qE0λ[1-e-λL]

     

  • (v0qB02λ)[2-e-2λL]

     

  • qE0λ[1+e-λL]

     

  • q[E0+v0B0λ][1-e-λL/2]

     

(1)

ΔK=WE+WM=WE+0

WE=∫0LqE dx

WE=∫0LqE0e-λx dx=qE0[e-λx-λ]0L

⇒WE=qE0λ[1-e-λL]

⇒ΔKE=qE0λ(1-e-λL)



Q 27 :

A current carrying circular loop of radius 2 cm with unit normal n^=k^+i^2 is placed in a magnetic field, B→=B0(3i^+2k^). If B0=4×10-3 T and current I=1002 A, the torque experienced by the loop is ______ Wb.A. (π=3.14)                 [2026]

  • 16×10-5k^

     

  • 5024×10-7k^

     

  • 5024×10-7i^

     

  • 5024×10-7j^

     

(4)

τ→=IA→×B→=Iπr2n^×B→

τ→=1002×3.14×(2×10-2)2[k^+i^2]×B0(3i^+2k^)

   =314×2×4×10-42×B0(3j^-2j^)

τ→=1256×10-4×4×10-3(j^) Wb·A

    =5024×10-7(j^) Wb·A



Q 28 :

1 μC charge moving with velocity v→=(i^-2j^+3k^) m/s in the region of magnetic field B→=(2i^+3j^-5k^) T. The magnitude of force acting on it is α×10-6 N. The value of α is ________.                     [2026]



(171)

F→=q(v→×B→)=1×10-6|i^j^k^1-2323-5|

F→=10-6[i^+11j^+7k^]⇒F→=171×10-6 N

∴ α=171



Q 29 :

The charged particle moving in a uniform magnetic field of (3i^+2j^) T has an acceleration (4i^-x2j^) m/s2. The value of x is ________.           [2026]



(12)

F→B∝(v→×B→)⇒F→B⊥B→

a→⊥B→⇒a→·B→=0

12-x=0⇒x=12



Q 30 :

A 5 mg particle carrying a charge of 5π×10-6 C is moving with velocity of (3i^+2k^)×10-2 m/s in a region having magnetic field B→=0.1k^ Wb/m2. It moves a distance of α meter along k^ when it completes 5 revolutions. The value of α is ________.            [2026]



(2)

T=2πmqB

Pitch p=VT=v(2πmqB)

α=5×p=5×v∥(2πmqB)

⇒α=5×0.02×2π×5×10-65π×10-6×0.1=2 m