मराठी
CBSEScience (English Medium) Class 12
प्रश्नपत्रिका३०५१
पाठ्यपुस्तक उत्तरे२५९९७
MCQ ऑनलाइन सराव परीक्षा४३
महत्वाचे प्रश्न आणि उत्तरे२७४३४
संकल्पना स्पष्टीकरण आणि व्हिडिओ७७९
टाइम टेबल२५
अभ्यासक्रम

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

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