Moving Charges & Magnetism (Part 2)
1. The Lorentz Force
In Part 1, we learned that a moving charge produces a magnetic field. Now, we study the reverse: What happens when a moving charge enters an external magnetic field?
$$\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})$$
Magnetic Lorentz Force: If only Magnetic Field is present ($\vec{E} = 0$), the force is:
$$F = qvB \sin \theta$$
Where:
• $q$: Magnitude of charge
• $v$: Velocity
• $B$: Magnetic Field strength
• $\theta$: Angle between $\vec{v}$ and $\vec{B}$
1.1 Special Cases (Viva/MCQ)
- Case 1: Stationary Charge ($v=0$)
$F = q(0)B \sin \theta = 0$.
Conclusion: Magnetic field exerts NO force on a stationary charge. - Case 2: Parallel Motion ($\theta = 0^\circ$ or $180^\circ$)
$F = qvB \sin 0^\circ = 0$.
Conclusion: Charge moving parallel to B-field experiences NO force. - Case 3: Perpendicular Motion ($\theta = 90^\circ$)
$F = qvB \sin 90^\circ = qvB$ (Maximum Force).
2. Direction of Force (Fleming's Left Hand Rule)
Since Force is a vector product ($\vec{F} = q \vec{v} \times \vec{B}$), its direction is perpendicular to both $\vec{v}$ and $\vec{B}$.
[attachment_0](attachment)• Forefinger: Direction of Magnetic Field ($\vec{B}$).
• Middle Finger: Direction of Conventional Current (or motion of +ve charge).
• Thumb: Points in the direction of Force ($\vec{F}$).
Moving Charges & Magnetism (Part 2)
1. Motion in Perpendicular Magnetic Field
When a charged particle enters a magnetic field perpendicularly ($\theta = 90^\circ$), the magnetic force acts as a Centripetal Force, making the particle move in a Circular Path.
[attachment_1](attachment)Derivation of Radius ($r$)
Centripetal Force = Magnetic Force
$\frac{mv^2}{r} = qvB$
2. Time Period ($T$) and Frequency ($\nu$)
Time taken to complete one revolution:
$T = \frac{2\pi r}{v}$. Substituting $r = \frac{mv}{qB}$:
Frequency: $$\nu = \frac{qB}{2\pi m}$$
Notice that Time Period ($T$) and Frequency ($\nu$) are independent of velocity ($v$) and radius ($r$).
3. Helical Motion ($\theta \neq 90^\circ$)
If velocity $\vec{v}$ makes an angle $\theta$ with $\vec{B}$, the path is a Helix.
• $v_{\parallel} = v \cos \theta$ (Responsible for pitch).
• $v_{\perp} = v \sin \theta$ (Responsible for circular motion).
$$r = \frac{m v_{\perp}}{qB} = \frac{mv \sin \theta}{qB}$$
Pitch of Helix ($p$):
$$p = v_{\parallel} \times T = (v \cos \theta) \times \frac{2\pi m}{qB}$$
Moving Charges & Magnetism (Part 2)
1. Principle & Construction
Cyclotron: A device used to accelerate charged particles (protons, deuterons, ions) to high energies.
2. Working & Theory
Resonance Condition
The frequency of the applied voltage must match the Cyclotron Frequency.
$$\nu_c = \frac{qB}{2\pi m}$$3. Maximum Kinetic Energy
$$K.E._{max} = \frac{q^2 B^2 R^2}{2m}$$
4. Limitations of Cyclotron (Important)
- It cannot accelerate neutral particles like Neutrons.
- For Electrons: Electrons cannot be accelerated because their mass is very small. They quickly acquire relativistic speed, causing an increase in mass ($m = \frac{m_0}{\sqrt{1-v^2/c^2}}$) and hence loss of the resonance condition.
Moving Charges & Magnetism (Part 2)
1. Force on a Current Carrying Conductor
$$F = B I l \sin \theta$$
2. Force Between Two Parallel Currents
Force per unit Length ($f$):
Nature of Force:
• Parallel Currents (Same direction): Attract each other.
• Anti-Parallel Currents (Opposite direction): Repel each other.
Question Bank (Part 2)
Section A: MCQs (1 Mark)
Q1. A proton and an alpha particle enter a uniform magnetic field perpendicularly with the same speed. The ratio of their radii ($r_p : r_\alpha$) is:
(a) 1:2 (b) 1:1 (c) 2:1 (d) 4:1
Ans: (a) 1:2
Q2. Two long parallel current-carrying conductors carrying currents in opposite directions will:
(a) Attract each other (b) Repel each other (c) Neither attract nor repel (d) Rotate
Ans: (b) Repel each other.
Section B: Numericals (3 Marks)
Q3. An electron moves with a speed of $2 \times 10^7$ m/s perpendicular to a magnetic field of $0.1$ T. Calculate the radius of the path.
Solution:
Formula: $r = \frac{mv}{qB}$
$r = \frac{9.1 \times 10^{-31} \times 2 \times 10^7}{1.6 \times 10^{-19} \times 0.1}$
Ans: 1.1 mm (approx).
Q4. A horizontal wire 0.1 m long carries a current of 5 A. What magnitude of magnetic field is required to support its weight? (Mass of wire = $3 \times 10^{-3}$ kg).
Solution:
Magnetic Force = Weight ($mg$)
$IlB = mg \Rightarrow B = \frac{mg}{Il}$
Ans: $0.0588$ Tesla.
Section D: Long Answer (5 Marks)
Q5. Derive an expression for the force per unit length between two long straight parallel conductors. Define 1 Ampere.
(See Vol 4).
Q6. Explain the principle, construction, and working of a Cyclotron. Derive the expression for cyclotron frequency.
(See Vol 3).
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