Magnetic Force on Charged Particle in Electric and Magnetic Fields – Force on a moving charge in uniform magnetic and electric: NEET MCQs with Answers

15 free questions on Magnetic Force on Charged Particle in Electric and Magnetic Fields – Force on a moving charge in uniform magnetic and electric for NEET (UG), starting with previous-year NEET and AIPMT questions. Try each one, then tap Show answer for the answer and a short explanation.

Syllabus covered: Force on a moving charge in uniform magnetic and electric fields

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Question 1

Previous-year question
A proton and an alpha particle both enter a region of uniform magnetic field B, moving at right angles to the field B. If the radius of the circular orbits for both the particles is equal, and the kinetic energy acquired by the proton is 1 MeV, the energy acquired by the alpha particle will be [AIPMT 2015]
  1. 0.5 MeV
  2. 1.5 MeV
  3. 1 MeV
  4. 4 MeV
Show answer
Answer: (C) 1 MeV
Explanation: For a particle of kinetic energy K, r = √(2mK)/qB. Setting r_p = r_α gives √(2m_pK_p)/e = √(2 × 4m_p K_α)/2e, so √(K_p) = √(K_α). Hence K_α = K_p = 1 MeV.

Question 2

Previous-year question
A deuteron of kinetic energy 50 keV describes a circular orbit of radius 0.5 m in a plane perpendicular to the magnetic field B. The kinetic energy of the proton that describes a circular orbit of radius 0.5 m in the same plane with the same magnetic field B is [AIPMT 1991]
  1. 25 keV
  2. 50 keV
  3. 200 keV
  4. 100 keV
Show answer
Answer: (D) 100 keV
Explanation: For a charged particle on a circular path, K = B²q²r²/2m. The deuteron and proton carry the same charge and here have the same r and B, so K ∝ 1/m. Since the deuteron's mass is twice the proton's, K_p/K_d = m_d/m_p = 2, giving K_p = 2 × 50 = 100 keV.

Question 3

Previous-year question
A uniform magnetic field acts at right angles to the direction of motion of electrons. As a result, the electrons move in a circular path of radius 2 cm. If the speed of the electrons is doubled, the radius of the circular path will be [AIPMT 1991]
  1. 2.0 cm
  2. 0.5 cm
  3. 4.0 cm
  4. 1.0 cm
Show answer
Answer: (C) 4.0 cm
Explanation: From qvB = mv²/r, the radius is r = mv/qB, so r ∝ v when m, q and B are fixed. Doubling the speed doubles the radius: r′ = 2 × 2 = 4 cm.

Question 4

Previous-year question
A charge moving with velocity v in the x-direction is subjected to a field of magnetic induction in the negative x-direction. As a result, the charge will [AIPMT 1993]
  1. remain unaffected
  2. start moving in a circular path in the YZ-plane
  3. be retarded along the X-axis
  4. move along a helical path around the X-axis
Show answer
Answer: (A) remain unaffected
Explanation: Here the velocity and the magnetic field are antiparallel, so θ = 180° and F = qvB sin 180° = 0. With no force acting, the charge continues undisturbed along its original straight line.

Question 5

Previous-year question
An electron enters a region where the magnetic field B and the electric field E are mutually perpendicular. Then it [AIPMT 1994]
  1. will always move in the direction of B
  2. will always move in the direction of E
  3. will always possess circular motion
  4. can go undeflected also
Show answer
Answer: (D) can go undeflected also
Explanation: With mutually perpendicular E and B, the electron experiences F = qE + q(v × B). If its speed happens to satisfy v = E/B, the two forces cancel exactly, the net force is zero, and the electron passes through undeflected. Since this is possible, none of the 'always' statements can be correct.

Question 6

Previous-year question
A 10 eV electron circulates in a plane at right angles to a uniform field of magnetic induction 10⁻⁴ Wb/m² (= 1.0 G). The orbital radius of the electron is [AIPMT 1996]
  1. 12 cm
  2. 16 cm
  3. 11 cm
  4. 18 cm
Show answer
Answer: (C) 11 cm
Explanation: From ½mv² = 10 eV = 10 × 1.6 × 10⁻¹⁹ J, v² = (2 × 1.6 × 10⁻¹⁸)/(9.1 × 10⁻³¹) = 3.52 × 10¹², so v = 1.88 × 10⁶ m/s. Then r = mv/qB = (9.1 × 10⁻³¹ × 1.88 × 10⁶)/(1.6 × 10⁻¹⁹ × 10⁻⁴) ≈ 0.11 m = 11 cm.

Question 7

Previous-year question
A beam of electrons moves with constant velocity in a region where an electric field of strength 20 V m⁻¹ and a magnetic field of 0.5 T act simultaneously at right angles to the direction of motion of the electrons. The velocity of the electrons must be [AIPMT 1996]
  1. 8 m/s
  2. 20 m/s
  3. 40 m/s
  4. 1/40 m/s
Show answer
Answer: (C) 40 m/s
Explanation: Constant velocity requires zero net force, so the electric and magnetic forces balance: qE = qvB. Hence v = E/B = 20/0.5 = 40 m/s.
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Question 8

Previous-year question
A positively charged particle moving due east enters a region of uniform magnetic field directed vertically upwards. The particle will [AIPMT 1997]
  1. continue to move due East
  2. move in a circular orbit with its speed unchanged
  3. move in a circular orbit with its speed increased
  4. get deflected vertically upwards
Show answer
Answer: (B) move in a circular orbit with its speed unchanged
Explanation: With v towards East and B vertically upwards, F = qv × B is horizontal and directed towards the South, i.e. perpendicular to v and lying in the horizontal plane. Since v remains perpendicular to B throughout, the particle executes uniform circular motion in the horizontal plane. The magnetic force does no work, so the speed is unchanged.

Question 9

Previous-year question
A proton moving with a velocity of 3 × 10⁵ m/s enters a magnetic field of 0.3 T at an angle of 30° with the field. The radius of curvature of its path will be (e/m for proton = 10⁸ C/kg) [AIPMT 2000]
  1. 2 cm
  2. 0.5 cm
  3. 0.02 cm
  4. 1.25 cm
Show answer
Answer: (B) 0.5 cm
Explanation: Only the velocity component perpendicular to B contributes to the circular part of the helical path. Hence r = mv sin θ/qB = v sin θ/[(e/m)B] = (3 × 10⁵ × 0.5)/(10⁸ × 0.3) = 1.5 × 10⁵/3 × 10⁷ = 5 × 10⁻³ m = 0.5 cm.

Question 10

Previous-year question
In a certain region of space, the electric field E and the magnetic field B are perpendicular to each other. An electron enters the region perpendicular to the directions of both B and E and moves undeflected. The velocity of the electron is [AIPMT 2001]
  1. |E| / |B|
  2. E × B
  3. |B| / |E|
  4. E · B
Show answer
Answer: (A) |E| / |B|
Explanation: The electron travels undeflected only when the electric and magnetic forces balance exactly: q|E| = q|v||B|. This gives |v| = |E| / |B|. This is the principle of the velocity selector.

Question 11

Previous-year question
A charged particle of charge q and mass m enters perpendicularly into a magnetic field B. If the kinetic energy of the particle is E, the frequency of rotation is [AIPMT 2001]
  1. qB/mπ
  2. qB/2πm
  3. qBE/2πm
  4. qB/2πE
Show answer
Answer: (B) qB/2πm
Explanation: Equating the magnetic force to the centripetal force, qvB = mv²/r, gives the angular frequency ω = v/r = qB/m. The frequency of rotation is therefore ν = ω/2π = qB/2πm — independent of the kinetic energy E.

Question 12

Previous-year question
A charge q moves in a region where both an electric field E and a magnetic field B exist. The force acting on it is [AIPMT 2002]
  1. q(v × B)
  2. qE + q(v × B)
  3. qB + q(B × v)
  4. qB + q(E × v)
Show answer
Answer: (B) qE + q(v × B)
Explanation: The total force on a charge in combined electric and magnetic fields is the Lorentz force, F = F_electric + F_magnetic = qE + q(v × B) = q[E + (v × B)].

Question 13

Previous-year question
A charged particle moves through a magnetic field in a direction perpendicular to it. Then the [AIPMT 2003]
  1. acceleration remains unchanged
  2. velocity remains unchanged
  3. speed of the particle remains unchanged
  4. direction of the particle remains unchanged
Show answer
Answer: (C) speed of the particle remains unchanged
Explanation: The magnetic force is always perpendicular to the velocity, so it does no work and cannot change the magnitude of the velocity. The direction of v changes continuously (so velocity and the direction of motion do change), and the acceleration, being always perpendicular to v, also changes direction. Only the speed remains constant.

Question 14

Previous-year question
An electron moves in a circular orbit with a uniform speed v. It produces a magnetic field B at the centre of the circle. The radius of the circle is proportional to [AIPMT 2005]
  1. B/v
  2. v/B
  3. √(v/B)
  4. √(B/v)
Show answer
Answer: (C) √(v/B)
Explanation: The orbiting electron is equivalent to a current i = e/T = ev/2πr. The field at the centre is B = μ₀i/2r = μ₀ev/4πr². Rearranging, r² = μ₀ev/4πB, so r ∝ √(v/B).

Question 15

Previous-year question
When a charged particle moving with velocity v is subjected to a magnetic field of induction B, the force acting on it is non-zero. This implies that [AIPMT 2006]
  1. the angle between v and B is necessarily 90°
  2. the angle between v and B can have any value other than 90°
  3. the angle between v and B can have any value other than zero and 180°
  4. the angle between v and B is either zero or 180°
Show answer
Answer: (C) the angle between v and B can have any value other than zero and 180°
Explanation: The magnitude of the magnetic force is F = qvB sin θ. This vanishes only when sin θ = 0, i.e. when θ = 0° or 180°. For the force to be non-zero, θ may take any value except these two — it need not be exactly 90°, which merely gives the maximum force.

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