Questions & explanations
1. The magnetic dipole moment of a current-carrying loop is defined as:
- The product of the current and the area of the loop, with direction perpendicular to the plane of the loop given by the right-hand rule
- The product of the current and the circumference of the loop, directed along the current
- The ratio of the magnetic field at the centre of the loop to the current flowing through it
- The product of the resistance of the loop and its cross-sectional area
Answer: The product of the current and the area of the loop, with direction perpendicular to the plane of the loop given by the right-hand rule
The magnetic dipole moment of a current loop is a vector quantity defined as m = N I A, where N is the number of turns, I is the current, and A is the area vector (magnitude equal to the area of the loop and direction normal to the plane, following the right-hand rule). Option (A) correctly captures this definition. Option (B) is incorrect because circumference is not involved. Option (C) incorrectly relates magnetic field and current. Option (D) incorrectly involves resistance and area.
2. Which of the following pairs correctly matches a soft magnetic material and a hard magnetic material with their respective applications?
- Soft iron – permanent magnets; Steel – transformer cores
- Soft iron – transformer cores; Steel – permanent magnets
- Nickel – electromagnet cores; Alnico – magnetic recording tapes
- Ferrite – permanent magnets; Soft iron – magnetic shielding only
Answer: Soft iron – transformer cores; Steel – permanent magnets
Soft magnetic materials (e.g., soft iron) have low coercivity and narrow hysteresis loop, making them suitable for transformer cores, electromagnets, etc. Hard magnetic materials (e.g., steel, alnico) have high coercivity and broad loop, making them suitable for permanent magnets. Option A reverses the applications; C and D contain inaccuracies (nickel is not typically used for electromagnet cores; soft iron also used in transformers, not just shielding).
3. A rectangular loop of area 0.05 m² carries a current of 2 A and has 10 turns. It is placed in a uniform magnetic field of 0.4 T with its plane parallel to the field. The magnitude of the torque acting on the loop is:
- A) 0 N m
- B) 0.04 N m
- C) 0.4 N m
- D) 0.2 N m
Answer: C) 0.4 N m
Torque on a current loop is τ = N I A B sinθ, where θ is the angle between the magnetic field and the normal to the loop. When the plane of the loop is parallel to the field, the normal is perpendicular to the field, so θ = 90° and sinθ = 1. Substituting N = 10, I = 2 A, A = 0.05 m², B = 0.4 T, we get τ = 10 × 2 × 0.05 × 0.4 = 10 × 2 × 0.02 = 0.4 N m. Option A would be if θ = 0° (plane perpendicular to field), B is 1/10 of the correct value, D is half.
4. Two long parallel conductors carry currents of 2 A and 3 A in opposite directions and are placed 10 cm apart. What is the force per unit length between them? (μ₀ = 4π×10⁻⁷ T m A⁻¹)
- A) 1.2×10⁻⁵ N/m
- B) 1.2×10⁻⁶ N/m
- C) 2.4×10⁻⁵ N/m
- D) 2.4×10⁻⁶ N/m
Answer: A) 1.2×10⁻⁵ N/m
Force per unit length between two parallel conductors is given by F/L = (μ₀ I₁ I₂)/(2π d). Substituting μ₀ = 4π×10⁻⁷, I₁ = 2 A, I₂ = 3 A, d = 0.1 m, we get F/L = (4π×10⁻⁷ × 2 × 3)/(2π × 0.1) = (24π×10⁻⁷)/(0.2π) = 120×10⁻⁷ = 1.2×10⁻⁵ N/m. Option B misses a factor of 10 (d in cm wrongly), C uses product of currents without μ₀? Actually 2.4×10⁻⁵ would correspond to double the correct value, D is 1/5 of correct.
5. A current-carrying conductor is placed in a uniform magnetic field. The magnetic field points into the plane of the page, and the current flows vertically downward (from top to bottom). Using Fleming's left-hand rule, the direction of the magnetic force acting on the conductor is:
- A) upward
- B) downward
- C) to the left
- D) to the right
Answer: D) to the right
Use Fleming's left-hand rule: forefinger along the field (into the page), middle finger along the conventional current (vertically downward). The thumb then points to the right. Equivalently, the force on the conductor is F = I L x B; with current direction down and B into the page, the cross product points to the right. So the magnetic force on the conductor is directed to the right.
6. When a diamagnetic material is placed in a non-uniform magnetic field, it tends to:
- Move from weaker to stronger magnetic field region
- Move from stronger to weaker magnetic field region
- Remain stationary regardless of field gradient
- Align its magnetic moments parallel to the field
Answer: Move from stronger to weaker magnetic field region
Diamagnetic materials develop a weak induced magnetic moment opposite to the applied field. In a non-uniform field, they experience a force towards the weaker field region (repelled from the strong field). Paramagnetic and ferromagnetic materials are attracted towards stronger field regions. The other options describe behavior of paramagnetic/ferromagnetic materials or are incorrect.
7. A charged particle enters a uniform magnetic field at an angle θ (0 < θ < 90°) to the field lines. The path of the particle will be:
- A) a straight line
- B) a circle
- C) a helix
- D) a parabola
Answer: C) a helix
When a charged particle enters a magnetic field at an angle other than 0° or 90°, its velocity has a component parallel to the field (which remains constant) and a perpendicular component (which causes circular motion). The combination results in a helical path. Straight line occurs when θ = 0°, circle when θ = 90°, and a parabola is characteristic of electric field effects.
8. In a B-H hysteresis loop for a ferromagnetic material, the area enclosed by the loop represents:
- The magnetic potential energy stored per unit volume
- The energy lost as heat per unit volume per cycle of magnetization
- The coercivity of the material
- The retentivity of the material
Answer: The energy lost as heat per unit volume per cycle of magnetization
The area of the B-H hysteresis loop corresponds to the energy dissipated as heat (hysteresis loss) per unit volume during one complete cycle of magnetization. This is due to the irreversible domain wall movements. Option about stored energy is incorrect because not all energy is recoverable; coercivity and retentivity are specific points on the loop, not the area.
9. Which of the following statements correctly distinguishes diamagnetic, paramagnetic, and ferromagnetic materials?
- Diamagnetic materials have negative susceptibility, paramagnetic have small positive susceptibility, ferromagnetic have large positive susceptibility
- Diamagnetic materials have zero susceptibility, paramagnetic have negative susceptibility, ferromagnetic have large positive susceptibility
- Diamagnetic materials have large positive susceptibility, paramagnetic have small negative susceptibility, ferromagnetic have zero susceptibility
- All three have positive susceptibility, but ferromagnetic materials have the highest value
Answer: Diamagnetic materials have negative susceptibility, paramagnetic have small positive susceptibility, ferromagnetic have large positive susceptibility
Diamagnetic materials (e.g., bismuth) have a small negative magnetic susceptibility (χ < 0). Paramagnetic materials (e.g., aluminium) have a small positive susceptibility (χ > 0, typically 10^-5 to 10^-3). Ferromagnetic materials (e.g., iron) have a very large positive susceptibility (χ >> 1). The other options incorrectly assign signs or magnitudes.
10. A proton (mass = 1.67×10⁻²⁷ kg, charge = 1.6×10⁻¹⁹ C) moves perpendicular to a uniform magnetic field of 0.5 T with a speed of 4×10⁶ m/s. What is the radius of its circular path?
- A) 0.0835 m
- B) 0.167 m
- C) 0.334 m
- D) 0.668 m
Answer: A) 0.0835 m
The radius of the circular path is given by r = mv/(qB). Substituting m = 1.67×10⁻²⁷ kg, v = 4×10⁶ m/s, q = 1.6×10⁻¹⁹ C, B = 0.5 T, we get r = (1.67×10⁻²⁷ × 4×10⁶) / (1.6×10⁻¹⁹ × 0.5) = (6.68×10⁻²¹) / (8×10⁻²⁰) = 0.0835 m. Option B would result from using 2× the radius (e.g., forgetting charge), C from doubling B, D from using half the charge.
11. Which of the following correctly represents the Biot-Savart law in vector form for a current element I d\vec{l}?
- d\vec{B} = \frac{\mu_0}{4\pi} \frac{I d\vec{l} \times \hat{r}}{r^2}
- d\vec{B} = \frac{\mu_0}{4\pi} \frac{I d\vec{l} \cdot \hat{r}}{r^2}
- d\vec{B} = \frac{\mu_0}{4\pi} \frac{I d\vec{l} \times \vec{r}}{r^3}
- d\vec{B} = \frac{\mu_0}{4\pi} \frac{I d\vec{l} \cdot \vec{r}}{r^3}
Answer: d\vec{B} = \frac{\mu_0}{4\pi} \frac{I d\vec{l} \times \hat{r}}{r^2}
The Biot-Savart law states that the magnetic field due to a current element is directly proportional to the cross product of the current element vector and the unit vector from the element to the point, inversely proportional to the square of the distance. Option A gives the exact vector form with the cross product and the unit vector.
12. Which of the following statements correctly compares the magnetic field pattern of a solenoid with that of a bar magnet?
- The magnetic field lines outside a solenoid are similar to those outside a bar magnet, and the solenoid has a north pole and a south pole.
- The magnetic field inside a solenoid is uniform, while inside a bar magnet it is zero.
- A solenoid produces a magnetic field only when a current flows, whereas a bar magnet always produces a field; other than that, their patterns are completely different.
- Both produce identical field patterns everywhere, including inside the solenoid and inside the magnet.
Answer: A solenoid produces a magnetic field only when a current flows, whereas a bar magnet always produces a field; other than that, their patterns are completely different.
A current-carrying solenoid behaves like a bar magnet, with similar external field lines (emerging from one end, entering the other) and identifiable north and south poles. The internal field of a solenoid is uniform, while inside a bar magnet the field is not zero but the lines continue; however, the external patterns are analogous.