Unit 13: Magnetic Effects of Current and Magnetism — Online MCQ Test
PHYSICS · NEET (All) · NEET
Practice Unit 13: Magnetic Effects of Current and Magnetism with a free chapter-wise online MCQ test.
This chapter covers: Biot - Savart law and its application to current carrying circular loop. Ampere's law and its applications to infinitely long current carrying straight wire and solenoid. Force on....
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Unit 13: Magnetic Effects of Current and Magnetism — Important Questions & Answers
A long solenoid of 50 cm length having 100 turns carries a current of 2.5 A. The magnetic field at the centre of the solenoid is (μ₀ = 4π x 10⁻⁷ T m A⁻¹):
- A. 3.14 x 10⁻⁴ T
- B. 6.28 x 10⁻⁵ T
- C. 3.14 x 10⁻⁵ T
- D. 6.28 x 10⁻⁴ T
Answer: D. 6.28 x 10⁻⁴ T
B = μ₀ n I, where n = N/L = 100 / 0.5 = 200 turns/m. B = (4π x 10⁻⁷) * 200 * 2.5 = 4π x 10⁻⁷ * 500 = 2000π x 10⁻⁷ = 2π x 10⁻⁴ = 6.28 x 10⁻⁴ T.
B = μ₀ n I, where n = N/L = 100 / 0.5 = 200 turns/m. B = (4π x 10⁻⁷) * 200 * 2.5 = 4π x 10⁻⁷ * 500 = 2000π x 10⁻⁷ = 2π x 10⁻⁴ = 6.28 x 10⁻⁴ T.
An electron (mass 9 × 10^-31 kg and charge 1.6 × 10^-19 C) moving with speed c/100 (c = speed of light) is injected into a magnetic field B of magnitude 9 × 10^-4 T perpendicular to its direction of motion. We wish to apply an uniform electric E together with the magnetic field so that the electron does not deflect from its path. Then (speed of light c = 3 × 10^8 ms^-1)
- A. E is perpendicular to B and its magnitude is 27 × 10^4 V m^-1
- B. E is perpendicular to B and its magnitude is 27 × 10^2 V m^-1
- C. E is parallel to B and its magnitude is 27 × 10^2 V m^-1
- D. E is parallel to B and its magnitude is 27 × 10^4 V m^-1
Answer: B. E is perpendicular to B and its magnitude is 27 × 10^2 V m^-1
For zero deflection, F_net = qE - qvB = 0, so E = vB. v = c/100 = 3*10^6 m/s. E = 3*10^6 * 9*10^-4 = 27*10^2 V/m. E must be perpendicular to B to balance the magnetic force.
For zero deflection, F_net = qE - qvB = 0, so E = vB. v = c/100 = 3*10^6 m/s. E = 3*10^6 * 9*10^-4 = 27*10^2 V/m. E must be perpendicular to B to balance the magnetic force.
A 2 amp current is flowing through two different small circular copper coils having radii ratio 1:2. The ratio of their respective magnetic moments will be
- A. 1:4
- B. 1:2
- C. 2:1
- D. 4:1
Answer: A. 1:4
Magnetic moment M = I * Area = I * pi * r^2. Since I is same, M is proportional to r^2. Ratio = (1)^2 : (2)^2 = 1:4.
Magnetic moment M = I * Area = I * pi * r^2. Since I is same, M is proportional to r^2. Ratio = (1)^2 : (2)^2 = 1:4.
A parallel plate capacitor is being charged such that the surface charge density is increasing at a constant rate. The magnetic field arising due to displacement current is:
- A. zero at all places
- B. constant between the plates and zero outside
- C. non-zero everywhere with maximum at the imaginary cylindrical surface connecting peripheries of the plates.
- D. zero between the plates and non-zero outside.
Answer: C. non-zero everywhere with maximum at the imaginary cylindrical surface connecting peripheries of the plates.
Displacement current exists between the plates. By Ampere-Maxwell law, magnetic field is generated. It varies linearly with radial distance r inside the capacitor and as 1/r outside, making it non-zero everywhere with a maximum at the rim.
Displacement current exists between the plates. By Ampere-Maxwell law, magnetic field is generated. It varies linearly with radial distance r inside the capacitor and as 1/r outside, making it non-zero everywhere with a maximum at the rim.
A parallel plate capacitor made of circular plates is being charged such that the surface charge density on its plates is increasing at a constant rate with time. The magnetic field arising due to displacement current is:
- A. zero at all places
- B. constant between the plates and zero outside the plates.
- C. non-zero everywhere with maximum at the imaginary cylindrical surface connecting peripheries of the plates.
- D. zero between the plates and non-zero outside.
Answer: C. non-zero everywhere with maximum at the imaginary cylindrical surface connecting peripheries of the plates.
For a charging parallel plate capacitor, the displacement current exists between the plates. By Ampere-Maxwell law, the magnetic field is symmetric and non-zero inside (between the plates) and zero outside due to circular symmetry.
For a charging parallel plate capacitor, the displacement current exists between the plates. By Ampere-Maxwell law, the magnetic field is symmetric and non-zero inside (between the plates) and zero outside due to circular symmetry.
The magnetic susceptibility is negative for:
- A. paramagnetic material only
- B. ferromagnetic material only
- C. paramagnetic and ferromagnetic materials
- D. diamagnetic material only
Answer: D. diamagnetic material only
Magnetic susceptibility (chi) is negative only for diamagnetic materials, as they are weakly repelled by magnetic fields.
Magnetic susceptibility (chi) is negative only for diamagnetic materials, as they are weakly repelled by magnetic fields.
A square loop ABCD carrying a current i is placed near and coplanar with a long straight conductor XY carrying a current I, the net force on the loop will be:
- A. (mu_0 * i * I) / (2 * pi)
- B. (2 * mu_0 * i * I * L) / (3 * pi)
- C. (mu_0 * i * I * L) / (2 * pi)
- D. (2 * mu_0 * i * I) / (3 * pi)
Answer: D. (2 * mu_0 * i * I) / (3 * pi)
The forces on the segments parallel to the wire are calculated using F = (mu_0 * I1 * I2 * L) / (2 * pi * d). The closer side experiences an attractive force and the farther side experiences a repulsive force. The net force is the difference between these two forces.
The forces on the segments parallel to the wire are calculated using F = (mu_0 * I1 * I2 * L) / (2 * pi * d). The closer side experiences an attractive force and the farther side experiences a repulsive force. The net force is the difference between these two forces.
A long straight wire of radius a carries a steady current I. The current is uniformly distributed over its cross-section. The ratio of the magnetic fields B and B' at radial distance a/2 and 2a respectively, from the axis of the wire is:
- A. 1/2
- B. 1
- C. 4
- D. 1/4
Answer: B. 1
Inside the wire, B = (μ₀Ir)/(2πa²). At r = a/2, B = μ₀I/(4πa). Outside the wire, B' = (μ₀I)/(2πr). At r = 2a, B' = μ₀I/(4πa). Thus, the ratio B/B' = 1:1.
Inside the wire, B = (μ₀Ir)/(2πa²). At r = a/2, B = μ₀I/(4πa). Outside the wire, B' = (μ₀I)/(2πr). At r = 2a, B' = μ₀I/(4πa). Thus, the ratio B/B' = 1:1.
In which of the following devices, the eddy current effect is not used?
- A. Electric heater
- B. Induction furnace
- C. Magnetic braking in train
- D. Electromagnet
Answer: A. Electric heater
Electric heaters operate on the principle of Joule heating (I²Rt), not eddy currents.
Electric heaters operate on the principle of Joule heating (I²Rt), not eddy currents.
The net magnetic flux through any closed surface is:
- A. Zero
- B. Positive
- C. Infinity
- D. Negative
Answer: A. Zero
According to Gauss's Law for magnetism, the magnetic monopoles do not exist, so net magnetic flux through any closed surface is always zero.
According to Gauss's Law for magnetism, the magnetic monopoles do not exist, so net magnetic flux through any closed surface is always zero.