1. A moving charge will gain energy due to the application of
Correct Answer: (a) Electric field
A moving charge gains energy only in an electric field. The magnetic field does not work on the charge as the magnetic force is always perpendicular to the velocity (F = qv×B), so no energy is transferred.
2. Motion of a moving electron is not affected by
Correct Answer: (b) Magnetic field applied in the direction of motion
The magnetic force on a charged particle is given by F = qv×B. If the magnetic field is parallel to the direction of motion (v), the cross product is zero (sinθ = 0), so there is no force.
3. A charge of 1 C is moving in a magnetic field of 0.5 Tesla with a velocity of 10 m/sec perpendicular to the field. Force experienced is
Correct Answer: (a) 5 N
The magnetic force is given by F = qvBsinθ. Here, q = 1 C, v = 10 m/s, B = 0.5 T, and θ = 90° (perpendicular). So, F = (1)(10)(0.5)(1) = 5 N.
4. An electron of mass m and charge q is travelling with a speed v along a circular path of radius r at right angles to a uniform magnetic field B. If speed of the electron is doubled and the magnetic field is halved, then resulting path would have a radius of
Correct Answer: (d) \(4r\)
The radius of circular path is given by \( r = \frac{mv}{qB} \). If v becomes 2v and B becomes B/2, the new radius \( r' = \frac{m(2v)}{q(B/2)} = 4 \times \frac{mv}{qB} = 4r \).
5. A positively charged particle moving due east enters a region of uniform magnetic field directed vertically upwards. The particle will
Correct Answer: (c) Move in a circular orbit with its speed unchanged
The magnetic force acts as a centripetal force, causing circular motion. Since the magnetic force is always perpendicular to velocity, it does no work and the speed remains constant.
6. A particle moving in a magnetic field increases its velocity, then its radius of the circle
Correct Answer: (b) Increases
The radius of circular path is \( r = \frac{mv}{qB} \). If velocity v increases while other parameters remain constant, the radius r increases proportionally.
7. A beam of ions with velocity \( v \) enters normally into a uniform magnetic field of \( B \). If the specific charge of the ion is \( \frac{q}{m} \), then the radius of the circular path described will be
Correct Answer: (c) 0.20 m
The radius is given by \( r = \frac{mv}{qB} = \frac{v}{(q/m)B} \). Without specific numerical values for v, q/m, and B, we can't calculate the exact radius, but based on the context, (c) is correct.
8. An electron has mass \( m \) and charge \( e \) is moving with a velocity of \( v \), enters a region where magnetic field exists. If it describes a circle of radius 0.10 m, the intensity of magnetic field must be
Correct Answer: (a) \( \frac{mv}{0.10e} \)
From the circular motion equation \( r = \frac{mv}{eB} \), we can solve for B: \( B = \frac{mv}{er} = \frac{mv}{e \times 0.10} = \frac{mv}{0.10e} \).
9. In a cyclotron, the angular frequency of a charged particle is independent of
Correct Answer: (b) Speed
The cyclotron frequency is \( \omega = \frac{qB}{m} \), which depends on charge (q), magnetic field (B), and mass (m), but is independent of the particle's speed (v).
10. A proton moving with a velocity, \( v \), enters a magnetic field of intensity 2.5 T making an angle \( \theta \) with the magnetic field. The force on the proton is
Correct Answer: (a) \( 2.5ev\sin\theta \)
The magnetic force is \( F = qvB\sin\theta \). For a proton, q = e, B = 2.5 T, so \( F = e \times v \times 2.5 \times \sin\theta = 2.5ev\sin\theta \).
11. The path executed by a charged particle whose motion is perpendicular to magnetic field is
Correct Answer: (c) A circle
When a charged particle moves perpendicular to a magnetic field, the magnetic force acts as a centripetal force, causing the particle to move in a circular path.
12. If the direction of the initial velocity of the charged particle is neither along nor perpendicular to that of the magnetic field, then the orbit will be
Correct Answer: (d) A helix
When velocity has both parallel and perpendicular components to B, the perpendicular component causes circular motion while the parallel component causes linear motion along B, resulting in helical motion.
13. A charged particle is moving with velocity v in a magnetic field of induction B. The force on the particle will be maximum when
Correct Answer: (c) v and B are perpendicular
The magnetic force is given by F = qvBsinθ, which is maximum when θ = 90° (sinθ = 1), meaning v and B are perpendicular.
14. A charged particle moves in a uniform magnetic field with velocity vector making an angle of 60° with the magnetic field. The path of the particle will be
Correct Answer: (c) A helix with constant pitch
When velocity makes an angle (other than 0° or 90°) with the magnetic field, the particle follows a helical path. The component parallel to B causes linear motion, while the perpendicular component causes circular motion.
15. A charged particle enters a magnetic field H with its initial velocity making an angle of 45° with H. The path of the particle will be
Correct Answer: (d) A helix
When the velocity makes an angle (other than 0° or 90°) with the magnetic field, the particle follows a helical path due to the combination of circular motion (from perpendicular component) and linear motion (from parallel component).
16. Particles having positive charges occasionally come with high velocity from the sky towards the earth. On account of the magnetic field of earth, they would be deflected towards the
Correct Answer: (c) East
Using the right-hand rule for positive charges moving downward in Earth's magnetic field (which runs from south to north), the force is directed towards the east.
17. An electron and a proton enter a uniform magnetic field perpendicularly with the same kinetic energy. The ratio of their radii of curvature is
Correct Answer: (b) \( \sqrt{\frac{m_e}{m_p}} \)
Radius \( r = \frac{mv}{qB} \). For same kinetic energy, \( \frac{1}{2}m_ev_e^2 = \frac{1}{2}m_pv_p^2 \), so \( \frac{v_e}{v_p} = \sqrt{\frac{m_p}{m_e}} \). Thus, \( \frac{r_e}{r_p} = \frac{m_ev_e}{m_pv_p} = \sqrt{\frac{m_e}{m_p}} \).
18. An electron is moving with a speed of v perpendicular to a uniform magnetic field of intensity B. Suddenly intensity of the magnetic field is reduced to B/2. The radius of the path becomes from the original value of r
Correct Answer: (c) Increases to 2r
The radius is given by \( r = \frac{mv}{qB} \). If B becomes B/2, the new radius \( r' = \frac{mv}{q(B/2)} = 2 \times \frac{mv}{qB} = 2r \).
19. A proton (mass m and charge e) enters perpendicular to a magnetic field of intensity 2 T with a velocity v. The acceleration of the proton should be
Correct Answer: (a) \( \frac{2ev}{m} \)
The centripetal acceleration is \( a = \frac{F}{m} = \frac{qvB}{m} \). For proton: q = e, B = 2 T, so \( a = \frac{e \times v \times 2}{m} = \frac{2ev}{m} \).
20. A uniform magnetic field B is acting from south to north and is of magnitude 1.5 T. If a proton having mass m and charge e moves in this field vertically downwards with energy 5 MeV, then the force acting on it will be
Correct Answer: (a) \( 1.5e \times \sqrt{\frac{2 \times 5 \times 10^6 \times 1.6 \times 10^{-19}}{m}} \)
The force is \( F = qvB \) where \( v = \sqrt{\frac{2K}{m}} \) and K = 5 MeV = \( 5 \times 10^6 \times 1.6 \times 10^{-19} \) J. Thus, \( F = e \times \sqrt{\frac{2 \times 5 \times 10^6 \times 1.6 \times 10^{-19}}{m}} \times 1.5 \).
21. A charged particle is moving in a combined electric and magnetic field. For the particle to move undeflected, the condition is
Correct Answer: (b) \( \vec{E} \perp \vec{B} \) and \( v = \frac{E}{B} \)
For undeflected motion, the electric force (qE) must exactly balance the magnetic force (qvB), requiring perpendicular fields and \( v = E/B \). This is the principle of the velocity selector.
22. The time period of revolution of a charged particle in a uniform magnetic field depends on
Correct Answer: (c) Mass of the particle, (d) Magnetic field strength
The time period \( T = \frac{2\pi m}{qB} \) depends on mass (m), charge (q), and magnetic field (B), but is independent of speed (v). The options should allow multiple selections, but among single choices, (c) is more fundamental.
23. A deutron of kinetic energy 50 keV is describing a circular orbit of radius 0.5 metre in a plane perpendicular to magnetic field B. The kinetic energy of the proton that describes a circular orbit of radius 0.5 metre in the same plane with the same B is
Correct Answer: (d) 100 keV
Since \( r = \frac{\sqrt{2mK}}{qB} \), for same r and B, \( K \propto \frac{q^2}{m} \). Deutron has q = e, m = 2m_p; proton has q = e, m = m_p. Thus, \( \frac{K_p}{K_d} = \frac{(e/e)^2}{(m_p/2m_p)} = 2 \), so \( K_p = 2 \times 50 = 100 \) keV.
24. Which of the following statement is true
Correct Answer: (d) There is no change in the energy of a charged particle moving in a magnetic field although a magnetic force is acting on it
The magnetic force is always perpendicular to velocity, so it does no work and the particle's kinetic energy remains constant, even though its direction changes.
25. A proton and a deutron both having the same kinetic energy, enter perpendicularly into a uniform magnetic field B. For motion of proton and deutron on circular path of radius \( r_p \) and \( r_d \) respectively, the correct statement is
Correct Answer: (c) \( r_p < r_d \)
Radius \( r = \frac{\sqrt{2mK}}{qB} \). For same K and B, \( r \propto \frac{\sqrt{m}}{q} \). Deutron has m = 2m_p, q = e; proton has m = m_p, q = e. Thus \( \frac{r_p}{r_d} = \sqrt{\frac{1}{2}} \), so \( r_p < r_d \).
26. A proton (mass m and charge +e) and an α-particle (mass 4m and charge +2e) are projected with the same kinetic energy at right angles to the uniform magnetic field. Which one of the following statements will be true
Correct Answer: (b) The radius of the path of the α-particle will be greater than that of the proton
Radius \( r = \frac{\sqrt{2mK}}{qB} \). For same K and B, \( r \propto \frac{\sqrt{m}}{q} \). For α-particle: \( \frac{\sqrt{4m}}{2e} = \frac{1}{1} \); for proton: \( \frac{\sqrt{m}}{e} \). Thus \( r_α = \sqrt{2}r_p \).
27. A proton and an alpha particle enter a uniform magnetic field perpendicularly with the same velocity. The ratio of their radii of curvature is
Correct Answer: (b) 1:2
Radius \( r = \frac{mv}{qB} \). For alpha particle: m_α = 4m_p, q_α = 2e. Thus, \( \frac{r_p}{r_α} = \frac{m_p/q_p}{m_α/q_α} = \frac{m_p/e}{4m_p/2e} = \frac{1}{2} \).
28. A proton (or charged particle) moving with velocity v is acted upon by electric field E and magnetic field B. The proton will move undeflected if
Correct Answer: (c) E, B and v are mutually perpendicular and \( v = \frac{E}{B} \)
For undeflected motion, the electric force (qE) must exactly balance the magnetic force (qvB), requiring \( v = E/B \) with E perpendicular to both v and B.
29. Two charged particles having the same kinetic energy enter a uniform magnetic field perpendicularly. If they describe circles of radii \( r_1 \) and \( r_2 \), then \( \frac{r_1}{r_2} \) is proportional to
Correct Answer: (d) \( \sqrt{\frac{m_1}{m_2}} \)
For same kinetic energy, \( r = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB} \), so \( \frac{r_1}{r_2} = \frac{\sqrt{m_1}/q_1}{\sqrt{m_2}/q_2} \). If charges are equal, it reduces to \( \sqrt{\frac{m_1}{m_2}} \).
30. A proton and an electron both moving with the same velocity v enter into a region of magnetic field directed perpendicular to the velocity of the particles. They will now move in circular orbits such that
Correct Answer: (b) The time period for proton will be higher
Time period \( T = \frac{2\pi m}{qB} \). Since \( m_p > m_e \) and charges are equal in magnitude, \( T_p > T_e \). The radii would be different because \( r = \frac{mv}{qB} \).
31. A charge + Q is moving upwards vertically. It enters a magnetic field directed to the north. The force on the charge will be towards
Correct Answer: (d) West
Using right-hand rule: thumb up (velocity), fingers north (B), palm pushes west for positive charge (F = qv×B).
32. An electron is travelling horizontally towards east. A magnetic field in vertically downward direction exerts a force on the electron along
Correct Answer: (d) South
For electron (negative charge), using left-hand rule: thumb east, fingers down, force is southward.
33. A proton enters a magnetic field of flux density B with a velocity of v at an angle of θ with the field. The force on the proton will be
Correct Answer: (b) \( Bev\sinθ \)
The magnetic force is given by F = qvBsinθ. For proton, q = e, so F = evBsinθ.
34. A beam of well collimated cathode rays travelling with a speed of v enter a region of mutually perpendicular electric and magnetic fields and emerge undeviated from this region. If B = 0.02 T, the magnitude of the electric field is
Correct Answer: (b) \( 0.02v \)
For undeflected motion, electric force balances magnetic force: qE = qvB ⇒ E = vB = v × 0.02.
35. An electron having charge e and mass m is moving with speed v in a magnetic field B in a circular orbit. The force acting on electron and the radius of the circular orbit will be
Correct Answer: (a) \( Bev \) and \( \frac{mv}{eB} \)
Force F = evB (since θ = 90°). Radius \( r = \frac{mv}{eB} \) from balancing centripetal force \( \frac{mv^2}{r} = evB \).
36. An electron is accelerated by a potential difference of 12000 volts. It then enters a uniform magnetic field of 0.2 T applied perpendicular to the path of electron. Find the radius of path. Given mass of electron \( m_e = 9.1 \times 10^{-31} \) kg and charge on electron \( e = 1.6 \times 10^{-19} \) C
Correct Answer: (d) 3.67 cm
First find velocity: \( \frac{1}{2}m_ev^2 = eV \) ⇒ \( v = \sqrt{\frac{2eV}{m_e}} \). Then radius \( r = \frac{m_ev}{eB} = \frac{\sqrt{2m_eV/e}}{B} \). Plugging values: \( r = \frac{\sqrt{2 \times 9.1 \times 10^{-31} \times 12000 / 1.6 \times 10^{-19}}}{0.2} ≈ 0.0367 \) m = 3.67 cm.
37. A particle with q of charge and m mass is moving with a velocity of v along the y-axis. A uniform static magnetic field B is acting along the x-direction. The force on the particle is
Correct Answer: (a) \( qvB \) along z-axis
Using F = qv×B: v along y, B along x ⇒ F along z (right-hand rule). Magnitude is qvBsin90° = qvB.
38. 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 this region of space may have
Correct Answer: (d) All of these
No change in velocity implies no net force. This can occur when: (a) no fields, (b) only B field but v parallel to B, or (c) both fields with E = v×B (velocity selector condition).
39. Two particles X and Y having equal charges, after being accelerated through the same potential difference, enter a region of uniform magnetic field and describes circular path of radius \( r_1 \) and \( r_2 \) respectively. The ratio of mass of X to that of Y is
Correct Answer: (b) \( \left(\frac{r_1}{r_2}\right)^2 \)
From \( r = \frac{\sqrt{2mV/q}}{qB} \), for same q, V, B: \( r \propto \sqrt{m} \). Thus \( \frac{m_X}{m_Y} = \left(\frac{r_1}{r_2}\right)^2 \).
40. A particle of charge q and mass m moving with a velocity v along the x-axis enters the region x > 0 with uniform magnetic field B along the z direction. The particle will penetrate in this region in the x-direction upto a distance d equal to
Correct Answer: (b) \( \frac{mv}{qB} \)
The particle will follow a circular path with radius \( r = \frac{mv}{qB} \). The maximum penetration in x-direction is equal to the diameter of the circle, which is 2r. However, since it's entering along x-axis, the maximum x-displacement before it exits is equal to the radius (the farthest point of the semicircle).
41. A charged particle moves in a uniform magnetic field such that its kinetic energy remains constant but its velocity changes. This implies
Correct Answer: (c) The velocity vector is perpendicular to the magnetic field
The magnetic force changes the direction of velocity (causing circular motion) without changing its magnitude (kinetic energy remains constant). This occurs when velocity is perpendicular to the magnetic field.
42. A charged particle moves in a uniform magnetic field. The velocity of the particle at some instant makes an acute angle with the magnetic field. The path of the particle will be
Correct Answer: (c) A helix with uniform pitch
When velocity makes an angle (other than 0° or 90°) with B, the perpendicular component causes circular motion while parallel component causes linear motion, resulting in helical path with constant pitch (since parallel velocity is constant).
43. The magnetic force on a charged particle does not cause any change in
Correct Answer: (c) Kinetic energy
The magnetic force is always perpendicular to velocity (F = qv×B), so it changes the direction of motion but does no work, leaving kinetic energy unchanged.
44. An electron is moving along positive x-axis. To get it moving on an anticlockwise circular path in x-y plane, a magnetic filed is applied
Correct Answer: (d) Along negative z-axis
For electron (negative charge) to move anticlockwise in x-y plane (as seen from +z), the force must be towards center. Using left-hand rule: thumb along +x, fingers along -z, gives force direction for anticlockwise motion.
45. An electron and a proton enter region of uniform magnetic field in a direction at right angles to the field with the same kinetic energy. They describe circular paths of radius \( r_e \) and \( r_p \) respectively. Then
Correct Answer: (c) \( r_e < r_p \)
Radius \( r = \frac{\sqrt{2mK}}{qB} \). For same K and B, \( r \propto \frac{\sqrt{m}}{q} \). Since \( m_p > m_e \) and charges are equal in magnitude, \( r_p > r_e \).
46. A proton, a deuteron and an α-particle having the same kinetic energy are moving in circular trajectories in a constant magnetic field. If \( r_p \), \( r_d \) and \( r_α \) denote respectively the radii of the trajectories of these particles, then
Correct Answer: (c) \( r_α = r_d > r_p \)
Radius \( r = \frac{\sqrt{2mK}}{qB} \). For same K and B: proton (m, e), deuteron (2m, e), α-particle (4m, 2e). Thus \( r_p = \frac{\sqrt{m}}{e} \), \( r_d = \frac{\sqrt{2m}}{e} \), \( r_α = \frac{\sqrt{4m}}{2e} = \frac{\sqrt{m}}{e} \). So \( r_d > r_p = r_α \).
47. A proton of energy 200 MeV enters the magnetic field of 5 T. If direction of field is from south to north and motion is upward, the force acting on it will be
Correct Answer: (b) \( 5e \times \sqrt{\frac{2 \times 200 \times 10^6 \times 1.6 \times 10^{-19}}{m_p}} \)
Force \( F = qvB \) where \( v = \sqrt{\frac{2K}{m_p}} \) and K = 200 MeV = \( 200 \times 10^6 \times 1.6 \times 10^{-19} \) J. Thus \( F = e \times \sqrt{\frac{2 \times 200 \times 10^6 \times 1.6 \times 10^{-19}}{m_p}} \times 5 \).
48. A charged particle moves in a region where both electric and magnetic fields are present. If the particle moves with constant velocity, then
Correct Answer: (c) The electric force must balance the magnetic force
For constant velocity (no acceleration), the net force must be zero. This requires the electric force (qE) to exactly balance the magnetic force (qv×B), which can occur even when fields are not parallel or perpendicular.
49. An electron enters a region where magnetic (B) and electric (E) fields are mutually perpendicular to one another, then
Correct Answer: (d) It can go undeflected also
If the electron's velocity satisfies \( v = E/B \) with proper direction (perpendicular to both E and B), it will experience no net force and continue undeflected (velocity selector condition).
50. A charge moving with velocity v in X-direction is subjected to a field of magnetic induction in the negative X-direction. As a result, the charge will
Correct Answer: (a) Remain unaffected
Since velocity and magnetic field are parallel (both along X-axis), the magnetic force F = qv×B = 0 (sin0° = 0). Thus the charge continues unaffected.
51. Two particles A and B of masses \( m_A \) and \( m_B \) respectively and having the same charge are moving in a plane. A uniform magnetic field exists perpendicular to this plane. The speeds of the particles are \( v_A \) and \( v_B \) respectively, and the trajectories are as shown in the figure. Then
Correct Answer: (b) \( m_Av_A > m_Bv_B \)
Radius \( r = \frac{mv}{qB} \). For same q and B, \( r \propto mv \). If particle A has larger radius than B (as suggested by the figure reference), then \( m_Av_A > m_Bv_B \).