Problem: An electron moving with a speed of 5.0 × 10⁶ m/s enters a uniform magnetic field of 0.2 T at an angle of 30° with the field direction. Calculate the magnitude of the magnetic force acting on the electron. (Charge of electron e = 1.6 × 10⁻¹⁹ C)

Solution 2 Marks

Given:
$v = 5.0 \times 10^6\, \text{m/s},\; B = 0.2\, \text{T},\; \theta = 30^\circ,\; q = -1.6 \times 10^{-19}\, \text{C}$

Apply Lorentz force formula:
Magnitude of magnetic force: $F = |q| v B \sin\theta$
$F = (1.6 \times 10^{-19}) \times (5.0 \times 10^6) \times 0.2 \times \sin 30^\circ$
$F = (1.6 \times 10^{-19}) \times (5.0 \times 10^6) \times 0.2 \times 0.5$
$F = 8.0 \times 10^{-14}\, \text{N}$

Direction:
Using right-hand rule (for positive charge) and reversing for electron (negative charge), force direction is perpendicular to both velocity and magnetic field vectors.

$\boxed{8.0 \times 10^{-14}\,\text{N}}$

2. Radius of Circular Path in Magnetic Field

Problem: A proton enters a uniform magnetic field of 0.5 T perpendicular to the field direction with a kinetic energy of 1.6 × 10⁻¹³ J. Calculate the radius of the circular path described by the proton. (Mass of proton = 1.67 × 10⁻²⁷ kg, charge = 1.6 × 10⁻¹⁹ C)

Charged particle moving in circular path in uniform magnetic field — Proton moving in circular path perpendicular to uniform magnetic field B

Solution 3 Marks

Given:
$B = 0.5\, \text{T},\; K = 1.6 \times 10^{-13}\, \text{J},\; m = 1.67 \times 10^{-27}\, \text{kg},\; q = 1.6 \times 10^{-19}\, \text{C}$

Step 1: Find velocity from kinetic energy
$K = \frac{1}{2} m v^2 \Rightarrow v = \sqrt{\frac{2K}{m}}$
$v = \sqrt{\frac{2 \times 1.6 \times 10^{-13}}{1.67 \times 10^{-27}}} = \sqrt{1.916 \times 10^{14}} = 1.384 \times 10^7\, \text{m/s}$

Step 2: Apply radius formula
For perpendicular motion ( $\theta = 90^\circ$ ), magnetic force provides centripetal force:
$q v B = \frac{m v^2}{r} \Rightarrow r = \frac{m v}{q B}$
$r = \frac{(1.67 \times 10^{-27}) \times (1.384 \times 10^7)}{(1.6 \times 10^{-19}) \times 0.5} = \frac{2.311 \times 10^{-20}}{8 \times 10^{-20}} = 0.289\, \text{m}$

$\boxed{0.289\,\text{m} \text{ or } 28.9\,\text{cm}}$

3. Helical Motion of Charged Particle

Problem: An electron with speed 4.0 × 10⁶ m/s enters a uniform magnetic field of 0.3 T at an angle of 60° with the field direction. Calculate (a) the radius of the helical path, (b) the pitch of the helix, and (c) the time period of revolution. (mₑ = 9.1 × 10⁻³¹ kg, e = 1.6 × 10⁻¹⁹ C)

Solution 3 Marks

Given:
$v = 4.0 \times 10^6\, \text{m/s},\; B = 0.3\, \text{T},\; \theta = 60^\circ,\; m = 9.1 \times 10^{-31}\, \text{kg},\; q = 1.6 \times 10^{-19}\, \text{C}$

Step 1: Resolve velocity components
Parallel component: $v_{\parallel} = v \cos\theta = 4.0 \times 10^6 \times \cos 60^\circ = 2.0 \times 10^6\, \text{m/s}$
Perpendicular component: $v_{\perp} = v \sin\theta = 4.0 \times 10^6 \times \sin 60^\circ = 3.464 \times 10^6\, \text{m/s}$

Part (a): Radius of helix
$r = \frac{m v_{\perp}}{q B} = \frac{(9.1 \times 10^{-31}) \times (3.464 \times 10^6)}{(1.6 \times 10^{-19}) \times 0.3} = \frac{3.152 \times 10^{-24}}{4.8 \times 10^{-20}} = 6.57 \times 10^{-5}\, \text{m}$

Part (b): Pitch of helix
Time period: $T = \frac{2\pi m}{q B} = \frac{2\pi \times 9.1 \times 10^{-31}}{1.6 \times 10^{-19} \times 0.3} = 1.19 \times 10^{-10}\, \text{s}$
Pitch: $p = v_{\parallel} T = (2.0 \times 10^6) \times (1.19 \times 10^{-10}) = 2.38 \times 10^{-4}\, \text{m}$

$\boxed{\text{(a) } 6.57 \times 10^{-5}\,\text{m} \quad \text{(b) } 2.38 \times 10^{-4}\,\text{m} \quad \text{(c) } 1.19 \times 10^{-10}\,\text{s}}$

4. Biot-Savart Law: Field Due to Current Element

Problem: A current element of length 2 cm carrying a current of 10 A is placed along the x-axis with its center at the origin. Calculate the magnetic field at a point (0, 2 m, 0) using Biot-Savart law. (μ₀/4π = 10⁻⁷ Tm/A)

Solution 2 Marks

Given:
$dl = 2\, \text{cm} = 0.02\, \text{m},\; I = 10\, \text{A},\; r = 2\, \text{m},\; \theta = 90^\circ$ (angle between dl and r)

Apply Biot-Savart law:
Magnitude: $dB = \frac{\mu_0}{4\pi} \frac{I \, dl \, \sin\theta}{r^2}$
$dB = 10^{-7} \times \frac{10 \times 0.02 \times \sin 90^\circ}{2^2} = 10^{-7} \times \frac{0.2}{4} = 5 \times 10^{-9}\, \text{T}$

Direction:
Using right-hand rule: dl along +x, r along +y ⇒ dl × r along +z direction
Therefore, magnetic field is along positive z-axis.

$\boxed{5 \times 10^{-9}\,\text{T along +z direction}}$

5. Magnetic Field on Axis of Circular Loop

Problem: A circular coil of radius 10 cm has 50 turns and carries a current of 2 A. Calculate the magnetic field at a point on its axis at a distance of (a) 0 cm (center), and (b) 15 cm from the center of the coil.

Solution 3 Marks

Given:
$R = 0.1\, \text{m},\; N = 50,\; I = 2\, \text{A},\; \frac{\mu_0}{4\pi} = 10^{-7}\, \text{Tm/A}$

Formula for field on axis:
$B = \frac{\mu_0 N I R^2}{2(R^2 + x^2)^{3/2}}$

Part (a): At center (x = 0)
$B = \frac{\mu_0 N I}{2R} = \frac{4\pi \times 10^{-7} \times 50 \times 2}{2 \times 0.1} = \frac{4\pi \times 10^{-5}}{0.2} = 6.28 \times 10^{-4}\, \text{T}$

Part (b): At x = 15 cm = 0.15 m
$B = \frac{4\pi \times 10^{-7} \times 50 \times 2 \times (0.1)^2}{2[(0.1)^2 + (0.15)^2]^{3/2}} = \frac{4\pi \times 10^{-7} \times 100 \times 0.01}{2[0.01 + 0.0225]^{3/2}}$
$B = \frac{1.256 \times 10^{-7}}{2 \times (0.0325)^{3/2}} = \frac{1.256 \times 10^{-7}}{2 \times 0.00586} = \frac{1.256 \times 10^{-7}}{0.01172} = 1.07 \times 10^{-5}\, \text{T}$

$\boxed{\text{(a) } 6.28 \times 10^{-4}\,\text{T} \quad \text{(b) } 1.07 \times 10^{-5}\,\text{T}}$

6. Ampere’s Circuital Law: Field Due to Straight Wire

Problem: A long straight wire carries a current of 25 A. Calculate the magnitude of magnetic field at a perpendicular distance of (a) 5 cm, and (b) 10 cm from the wire. Also find the direction of the field at these points.

Magnetic field lines around straight current-carrying conductor — Concentric circular magnetic field lines around a straight current-carrying wire (right-hand thumb rule)

Solution 2 Marks

Given:
$I = 25\, \text{A},\; r_1 = 0.05\, \text{m},\; r_2 = 0.10\, \text{m}$

Apply Ampere’s circuital law:
For infinite straight wire: $B = \frac{\mu_0 I}{2\pi r}$

Part (a): At r = 5 cm
$B_1 = \frac{4\pi \times 10^{-7} \times 25}{2\pi \times 0.05} = \frac{2 \times 10^{-5}}{0.05} = 4 \times 10^{-4}\, \text{T} = 0.4\, \text{mT}$

Part (b): At r = 10 cm
$B_2 = \frac{4\pi \times 10^{-7} \times 25}{2\pi \times 0.10} = \frac{2 \times 10^{-5}}{0.10} = 2 \times 10^{-4}\, \text{T} = 0.2\, \text{mT}$

Direction:
Using right-hand thumb rule: Grasp wire with thumb pointing in current direction; fingers curl in direction of magnetic field (tangential to concentric circles).

$\boxed{\text{(a) } 0.4\,\text{mT} \quad \text{(b) } 0.2\,\text{mT}}$

7. Magnetic Field Inside a Solenoid

Problem: A solenoid 0.8 m long has 2000 turns and carries a current of 3.0 A. Calculate (a) the magnetic field inside the solenoid near its center, and (b) the magnetic moment of the solenoid if its cross-sectional area is 4 cm².

Solution 2 Marks

Given:
$l = 0.8\, \text{m},\; N = 2000,\; I = 3.0\, \text{A},\; A = 4 \times 10^{-4}\, \text{m}^2$

Part (a): Magnetic field inside solenoid
Turns per unit length: $n = \frac{N}{l} = \frac{2000}{0.8} = 2500\, \text{turns/m}$
$B = \mu_0 n I = 4\pi \times 10^{-7} \times 2500 \times 3.0 = 9.42 \times 10^{-3}\, \text{T} = 9.42\, \text{mT}$

Part (b): Magnetic moment of solenoid
Total magnetic moment: $m = N I A = 2000 \times 3.0 \times 4 \times 10^{-4} = 2.4\, \text{A·m}^2$
Direction: Along axis following right-hand rule (curl fingers in current direction, thumb gives m direction)

$\boxed{\text{(a) } 9.42\,\text{mT} \quad \text{(b) } 2.4\,\text{A·m}^2}$

8. Force Between Parallel Current-Carrying Conductors

Problem: Two long parallel conductors are 10 cm apart and carry currents of 15 A and 25 A respectively in the same direction. Calculate the force per unit length experienced by each conductor. Is the force attractive or repulsive?

Solution 2 Marks

Given:
$I_1 = 15\, \text{A},\; I_2 = 25\, \text{A},\; d = 0.10\, \text{m}$

Apply force formula for parallel conductors:
Force per unit length: $f = \frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}$
$f = \frac{4\pi \times 10^{-7} \times 15 \times 25}{2\pi \times 0.10} = \frac{2 \times 10^{-7} \times 375}{0.10} = \frac{7.5 \times 10^{-5}}{0.10} = 7.5 \times 10^{-4}\, \text{N/m}$

Nature of force:
Since currents are in the same direction, the force is attractive.
(Parallel currents attract, antiparallel currents repel)

Newton’s third law verification:
Force on conductor 1 due to 2 = Force on conductor 2 due to 1 = $7.5 \times 10^{-4}\, \text{N/m}$
Equal in magnitude, opposite in direction (attractive)

$\boxed{7.5 \times 10^{-4}\,\text{N/m (attractive)}}$

9. Torque on Current Loop in Uniform Magnetic Field

Problem: A rectangular coil of 100 turns, 8 cm × 5 cm, carries a current of 2 A. It is placed in a uniform magnetic field of 0.4 T such that its plane makes an angle of 30° with the field direction. Calculate the torque experienced by the coil.

Solution 3 Marks

Given:
$N = 100,\; l = 0.08\, \text{m},\; b = 0.05\, \text{m},\; I = 2\, \text{A},\; B = 0.4\, \text{T},\; \alpha = 30^\circ$
Area: $A = l \times b = 0.08 \times 0.05 = 4 \times 10^{-3}\, \text{m}^2$

Key concept:
Angle between magnetic field B and normal to coil plane = θ
Given: Plane makes 30° with field ⇒ Normal makes θ = 90° – 30° = 60° with field

Apply torque formula:
$\tau = N I A B \sin\theta = 100 \times 2 \times (4 \times 10^{-3}) \times 0.4 \times \sin 60^\circ$
$\tau = 100 \times 2 \times 4 \times 10^{-3} \times 0.4 \times \frac{\sqrt{3}}{2} = 0.32 \times \sqrt{3} = 0.554\, \text{N·m}$

Vector form:
Magnetic moment: $\vec{m} = N I \vec{A}$
Torque: $\vec{\tau} = \vec{m} \times \vec{B}$

$\boxed{0.554\,\text{N·m}}$

10. Moving Coil Galvanometer Sensitivity

Problem: A moving coil galvanometer has a coil of 50 turns, each of area 3 × 10⁻⁴ m², suspended in a radial magnetic field of 0.2 T. The torsional constant of the spring is 2 × 10⁻⁴ Nm/rad. Calculate (a) the current sensitivity, and (b) the voltage sensitivity of the galvanometer.

Moving coil galvanometer with radial magnetic field and soft iron core — Moving coil galvanometer with radial magnetic field ensuring constant torque at all deflection angles

Solution 3 Marks

Given:
$N = 50,\; A = 3 \times 10^{-4}\, \text{m}^2,\; B = 0.2\, \text{T},\; k = 2 \times 10^{-4}\, \text{Nm/rad}$

Part (a): Current sensitivity
Defined as deflection per unit current: $\frac{\phi}{I} = \frac{NAB}{k}$
$\frac{\phi}{I} = \frac{50 \times 3 \times 10^{-4} \times 0.2}{2 \times 10^{-4}} = \frac{3 \times 10^{-3}}{2 \times 10^{-4}} = 15\, \text{rad/A} = 15 \times \frac{180}{\pi} \approx 860\, \text{divisions/A}$
(Assuming 1 rad ≈ 57.3° and scale calibrated accordingly)

Part (b): Voltage sensitivity
First find coil resistance (assume typical value for calculation):
For galvanometer coil: $R \approx 50\, \Omega$ (typical value for moving coil instruments)
Voltage sensitivity: $\frac{\phi}{V} = \frac{\phi}{I} \times \frac{1}{R} = \frac{15}{50} = 0.3\, \text{rad/V}$

Important note:
• Current sensitivity depends on N, A, B and k
• Voltage sensitivity = Current sensitivity / Resistance
• Increasing N increases current sensitivity but also increases R, so voltage sensitivity may not increase proportionally

$\boxed{\text{(a) } 15\,\text{rad/A} \quad \text{(b) } 0.3\,\text{rad/V (for } R = 50\,\Omega)}$

Exam Tip: For numerical problems on Moving Charges and Magnetism:

• Always resolve velocity into components parallel (v∥) and perpendicular (v⊥) to B for helical motion
• For circular motion: centripetal force = magnetic force ⇒ mv²/r = qvB ⇒ r = mv/qB
• Magnetic force does NO WORK (always perpendicular to velocity) ⇒ kinetic energy remains constant
• For Biot-Savart law: dB ∝ I·dl·sinθ/r²; direction by right-hand rule for cross product
• Ampere’s law useful for symmetric current distributions (straight wire, solenoid, toroid)
• Parallel currents ATTRACT; antiparallel currents REPEL (opposite to electrostatic behavior)
• Torque on loop: τ = m × B where m = NIA (magnetic moment)

← Back to Theory Next: Magnetism & Matter →

10 Solved Numericals on Moving Charges and Magnetism

10 Solved Numericals on Moving Charges and Magnetism

1. Lorentz Force on Moving Charge

2. Radius of Circular Path in Magnetic Field

3. Helical Motion of Charged Particle

4. Biot-Savart Law: Field Due to Current Element

5. Magnetic Field on Axis of Circular Loop

6. Ampere’s Circuital Law: Field Due to Straight Wire

7. Magnetic Field Inside a Solenoid

8. Force Between Parallel Current-Carrying Conductors

9. Torque on Current Loop in Uniform Magnetic Field

10. Moving Coil Galvanometer Sensitivity

Editorial Team

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1. Lorentz Force on Moving Charge

2. Radius of Circular Path in Magnetic Field

3. Helical Motion of Charged Particle

4. Biot-Savart Law: Field Due to Current Element

5. Magnetic Field on Axis of Circular Loop

6. Ampere’s Circuital Law: Field Due to Straight Wire

7. Magnetic Field Inside a Solenoid

8. Force Between Parallel Current-Carrying Conductors

9. Torque on Current Loop in Uniform Magnetic Field

10. Moving Coil Galvanometer Sensitivity

Editorial Team

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