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CBSEScience (English Medium) Class 12
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अवधारणा स्पष्टीकरण और वीडियो७७९
समय सारणी२५
पाठ्यक्रम

Motion in a Magnetic Field

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Estimated time: 9 minutes
CBSE: Class 12

Magnetic Force Changes Direction, Not Speed

A key foundational principle governs all motion in a magnetic field:

The magnetic force is always perpendicular to the velocity of the particle.

Because force has no component along the direction of motion:

  • No work is done by the magnetic force
  • Kinetic energy remains constant
  • Speed does not change — only direction changes
Property Magnetic Force Electric Force
Work done on the particle Zero Non-zero
Changes speed? No Yes
Changes direction? Yes Can (if perpendicular)
Transfers energy? No Yes
CBSE: Class 12

Case Analysis: Path of a Charged Particle in Magnetic Field

The path depends entirely on the angle θ between velocity \[\vec v\] and magnetic field \[\vec B\]:

CBSE: Class 12

Case 1: Circular Motion (v ⊥ B)

Condition

Velocity \[\vec v\] is perpendicular to the magnetic field \[\vec B\] (θ = 90°).

Mechanism

The magnetic force F = qvB acts perpendicular to v at every instant, exactly like a centripetal force. The particle is continuously deflected without gaining or losing speed — resulting in uniform circular motion.

Derivation of Radius

Step 1: Write the magnetic force on the particle:

Fmagnetic = qvB

Step 2: This force provides the centripetal force:

Fcentripetal = \[\frac {mv^2}{r}\]

Step 3: Equate both expressions:

\[\frac {mv^2}{r}\] = qvB

Step 4: Solve for radius r:

r = \[\frac{mv}{qB}\]   ...(4.5)
Inference: Larger momentum mvmv → larger radius. Heavier or faster particles trace wider circles.
CBSE: Class 12

Case 2: Cyclotron Frequency and Time Period

Angular Frequency

Since v = ωr, substituting in equation (4.5):

\[\omega r = \frac{qBr}{m} \implies{\omega = \frac{qB}{m}}\] …[4.6(a)][1]

Frequency of Revolution

ν = \[\frac {qB}{2πm}\]

Time Period

T = \[\frac {2πm}{qB}\] = \[\frac {1}{ν}\]
Critical Insight: Both ω, ν, and T are completely independent of the particle's speed and radius. This is the cyclotron principle — the basis of particle accelerators.
CBSE: Class 12

Case 3: Helical Motion (v at angle θ to B)

Condition

Velocity \[\vec v\] has both a perpendicular component and a parallel component to \[\vec B\] (0° < θ < 90°).

Resolution of Velocity

Resolve \[\vec v\] into two components:

Component Symbol Direction Effect
Perpendicular to B v⊥ = v sin⁡ θ ⊥ to B Causes circular motion
Parallel to B v∥ = v cos ⁡θ ∥ to B Causes linear motion (unchanged)

Resultant Path: Helix

The combination of circular + linear motion produces a helical path along the direction of B.

Pitch of the Helix

Pitch (p) = distance traveled along \[\vec B\] in one complete revolution:

\[{p = v_\parallel \cdot T = \frac{2\pi m\, v_\parallel}{qB}}\] …[4.6(b)][1]
CBSE: Class 12

Example

Example 4.3: An electron (mass 9 × 10−31 kg, charge 1.6 × 10−19 C) moves at 3 × 107 m/s in a magnetic field of 6 × 10−4 T perpendicular to its velocity. Find: (i) radius, (ii) frequency, (iii) kinetic energy in keV.

Solution:

(i) Radius:

r = \[\frac{mv}{qB}=\frac{9\times10^{-31}\times3\times10^{7}}{1.6\times10^{-19}\times6\times10^{-4}}\]​ = 0.28 m = 28 cm

(ii) Frequency:

ν = \[\frac{v}{2\pi r}=\frac{3\times10^7}{2\pi\times0.28}\]​ ≈ 17 MHz

(iii) Kinetic Energy:

KE = \[\frac {1}{2}\]mv2 = \[\frac {1}{2}\] × 9 × 10−31 × (3 × 107)2 ≈ 4 × 10−16 J = 2.5 k

Related QuestionsVIEW ALL [26]

Protons and singly ionized atoms of U235 and U238 are passed in turn (which means one after the other and not at the same time) through a velocity selector and then enter a uniform magnetic field. The protons describe semicircles of radius 10 mm. The separation between the ions of U235 and U238 after describing the semicircle is given by ______.

A conductor ABOCD moves along its bisector with a velocity 1 m/s through a perpendicular magnetic field of 1 wb/m2, as shown in figure. If all the four sides are 1 m length each, then the induced emf between A and Din approx is ______V.

Two proton beams going in the same direction repel each other whereas two wires carrying currents in the same direction attract each other. Explain.

Consider the situation shown in figure. The wires P1Q1 and P2Q2 are made to slide on the rails with the same speed 5 cm s−1. Suppose the 19 Ω resistor is disconnected. Find the current through P2Q2 if (a) both the wires move towards right and (b) if P1Q1 moves towards left but P2Q2 moves towards right.

Two identical coils P and Q each of radius R are lying in perpendicular planes such that they have a common centre. Find the magnitude and direction of the magnetic field at the common centre of the two coils, if they carry currents equal to I and \[\sqrt{3}\] I respectively.

The motion of copper plate is damped when it is allowed to oscillate between the two poles of a magnet. What is the cause of this damping?

Consider the following statements and select the incorrect statement(s).
  1. The presence of a large magnetic flux through a coil maintains a current in the coil if the circuit is continuous.
  2. A coil of a metal wire kept stationary in a non– uniform magnetic field has an e.m.f induced in it.
  3. A charged particle enters a region of uniform magnetic field at an angle of 85° to the magnetic lines of force, the path of the particle is a circle.
  4. There is no change in the energy of a charged particle moving in a magnetic field although a magnetic force is acting on it.

Find the condition under which the charged particles moving with different speeds in the presence of electric and magnetic field vectors can be used to select charged particles of a particular speed.

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