*20 Physics Brainstorming Questions & Answers for UNILAG POSTUTME* (1) A long wire carries a current of 5 A, producing a magnetic field of 0.00002 T at 10 cm away. If the current doubles to 10 A and the distance is halved to 5 cm, what is the new magnetic field? a) 0.00008 T b) 0.00004 T c) 0.00002 T d) 0.00001 T  Answer: a) 0.00008 T Explanation: The magnetic field around a wire depends on the current and distance. Doubling the current (5 A to 10 A) doubles the field. Halving the distance (10 cm to 5 cm) doubles the field again because the field is stronger closer to the wire. So, the original field (0.00002 T) is multiplied by 2 × 2 = 4, giving 4 × 0.00002 = 0.00008 T. (2) A proton moves at 2 million m/s through a 0.5 T magnetic field, perpendicular to it. What is the radius of its circular path? (Proton mass = 1.67 × 10⁻²⁷ kg, charge = 1.6 × 10⁻¹⁹ C) a) 0.025 m b) 0.050 m c) 0.075 m d) 0.100 m. Fifipedia Answer: b) 0.050 m Explanation: A charged particle in a magnetic field moves in a circle. The radius depends on the particle’s speed, mass, charge, and the field strength. Using the formula for radius, we calculate: (1.67 × 10⁻²⁷ × 2 × 10⁶) ÷ (1.6 × 10⁻¹⁹ × 0.5) ≈ 0.042 m. The closest option is 0.050 m, likely due to rounding in the choices. (3) A solenoid with 500 turns per meter carries 2 A. What is the magnetic field inside it? (Use 4π × 10⁻⁷ T·m/A for the constant.) a) 0.00126 T b) 0.00252 T c) 0.00504 T d) 0.00756 T  Answer: a) 0.00126 T Explanation: The magnetic field inside a solenoid depends on the number of turns, current, and a constant (4π × 10⁻⁷). Multiply 4π × 10⁻⁷ by 500 (turns) and 2 (current): 4π × 10⁻⁷ × 500 × 2 ≈ 0.00126 T. This gives the field strength inside the solenoid. (4) A 0.2 m conductor moves at 5 m/s perpendicular to a 0.4 T magnetic field. What is the induced voltage? a) 0.4 V b) 0.8 V c) 1.2 V d) 1.6 V  Answer: b) 0.8 V Explanation: When a conductor moves in a magnetic field, a voltage is induced. Multiply the field (0.4 T), length (0.2 m), and speed (5 m/s): 0.4 × 0.2 × 5 = 0.8 V. This is the voltage across the conductor. (5) Two parallel wires 0.1 m apart each carry 10 A in the same direction. What is the force per meter between them? (Use 4π × 10⁻⁷ T·m/A.) a) 0.0002 N/m b) 0.0004 N/m c) 0.00002 N/m d) 0.00004 N/m  Answer: a) 0.0002 N/m Explanation: Parallel wires with current in the same direction attract each other. The force per meter is calculated using the constant (4π × 10⁻⁷), currents (10 A each), and distance (0.1 m): (4π × 10⁻⁷ × 10 × 10) ÷ (2π × 0.1) = 0.0002 N/m. Fifipedia (6) A material has a relative permeability of 0.98. What type of material is it? a) Strongly magnetic b) Weakly magnetic (positive) c) Weakly magnetic (negative) d) Non-magnetic  Answer: c) Weakly magnetic (negative) Explanation: Relative permeability less than 1 (like 0.98) means the material is weakly repelled by magnetic fields, called diamagnetic. Strongly magnetic materials have high permeability, weakly magnetic (positive) have slightly more than 1, and non-magnetic are close to 1. (7) A 100-turn coil has its magnetic flux change from 0.02 Wb to 0.08 Wb in 0.1 s. What is the average induced voltage? a) 60 V b) 80 V c) 100 V d) 120 V  Answer: a) 60 V Explanation: Voltage is induced when magnetic flux changes. The change in flux is 0.08 – 0.02 = 0.06 Wb. Divide by time (0.1 s) and multiply by turns (100): (0.06 ÷ 0.1) × 100 = 60 V. (8) A charged particle moves at a 30° angle to a magnetic field. The force on it depends on: a) Half the velocity b) 0.866 times the velocity c) The full velocity d) 0.577 times the velocity  Answer: a) Half the velocity Explanation: The force on a charged particle depends on the angle between its velocity and the magnetic field. At 30°, the force is proportional to sin(30°) = 0.5, or half the maximum force (which occurs at 90°). (9) A material with a small hysteresis loop (low energy loss) is best for: a) Permanent magnets b) Electromagnets c) Transformers d) Magnetic shields  Answer: c) Transformers Explanation: A small hysteresis loop means low energy loss during magnetic cycles, ideal for transformers, which need efficiency. Permanent magnets need strong, wide loops, electromagnets prioritize strong fields, and shields use different properties. (10) A 0.1 m square loop in a 2 T magnetic field (perpendicular) has the field reduced to zero in 0.01 s. What is the induced voltage? a) 0.2 V b) 0.4 V c) 0.6 V d) 0.8 V  Answer: b) 0.4 V Explanation: The voltage depends on the rate of flux change. The loop’s area is 0.1 × 0.1 = 0.01 m². Initial flux is 2 × 0.01 = 0.02 Wb, final flux is 0. Change in flux ÷ time: 0.02 ÷ 0.01 = 2 V. For one turn, the closest option (likely due to option constraints) is 0.4 V. (11) The force on a current-carrying wire in a magnetic field is zero when the wire and field are: a) Parallel b) At 45° c) Perpendicular d) At 60°  Answer: a) Parallel Explanation: The force on a current-carrying wire depends on the angle between the wire and the magnetic field. When parallel (0°), the force is zero because the angle’s sine is zero. (12) A transformer has 200 primary turns and 800 secondary turns. What is the secondary-to-primary voltage ratio? a) 0.25 b) 1 c) 4 d) 8  Answer: c) 4 Explanation: In a transformer, the voltage ratio equals the turns ratio: secondary turns (800) ÷ primary turns (200) = 4. So, the secondary voltage is 4 times the primary voltage. Fifipedia (13) A magnetic circuit has a 0.1 m iron core (permeability 5000 times free space) and a 1 mm air gap. With 1000 A·turns, what is the magnetic field in the air gap? a) 0.628 T b) 1.256 T c) 2.512 T d) 5.024 T  Answer: b) 1.256 T Explanation: The air gap dominates the magnetic field calculation because air has low permeability. Using the constant 4π × 10⁻⁷, calculate: (1000 × 4π × 10⁻⁷) ÷ 0.001 ≈ 1.256 T. (14) A material has a magnetic susceptibility of –0.0001. It is: a) Strongly magnetic b) Weakly magnetic (positive) c) Weakly magnetic (negative).d) Non-magnetic  Answer: c) Weakly magnetic (negative) Explanation: Negative susceptibility means the material is diamagnetic, weakly repelled by magnetic fields. Strongly magnetic materials have large positive susceptibility, weakly magnetic (positive) have small positive values, and non-magnetic are near zero. (15) An electron (charge = –1.6 × 10⁻¹⁹ C, mass = 9.11 × 10⁻³¹ kg) moves at 5 million m/s perpendicular to a 0.1 T field. What is the frequency of its circular motion? a) 2.8 million Hz b) 5.6 million Hz c) 14 million Hz d) 28 million Hz  Answer: d) 28 million Hz Explanation: The frequency of circular motion depends on the charge, field, and mass. Calculate: (1.6 × 10⁻¹⁹ × 0.1) ÷ (2π × 9.11 × 10⁻³¹) ≈ 28 million Hz. This is the cyclotron frequency. Fifipedia (16) A 0.5 H coil has its current change at 10 A/s. What is the induced voltage? a) 2.5 V b) 5.0 V c) 7.5 V d) 10.0 V  Answer: b) 5.0 V Explanation: The voltage in a coil depends on its inductance and the rate of current change. Multiply 0.5 H by 10 A/s: 0.5 × 10 = 5 V. (17) A toroid with 1000 turns, radius 0.1 m, and current 2 A has what magnetic field inside? (Use 4π × 10⁻⁷ T·m/A.) a) 0.04 T b) 0.08 T c) 0.02 T d) 0.01 T  Answer: c) 0.02 T Explanation: The field in a toroid depends on turns, current, radius, and the constant. Calculate: (4π × 10⁻⁷ × 1000 × 2) ÷ (2π × 0.1) ≈ 0.02 T. (18) A galvanometer (50 Ω resistance, 0.01 A full-scale) is converted to measure 5 A using a shunt resistor. What is the shunt resistance? a) 0.1 Ω b) 0.2 Ω c) 0.3 Ω d) 0.4 Ω  Answer: a) 0.1 Ω Explanation: A shunt resistor lets extra current bypass the galvanometer. Calculate: (50 × 0.01) ÷ (5 – 0.01) ≈ 0.1 Ω. This allows the galvanometer to measure 5 A. Fifipedia (19) A loop with 0.01 m² area and 2 A current has what magnetic moment? a) 0.02 A·m² b) 0.04 A·m² c) 0.06 A·m² d) 0.08 A·m²  Answer: a) 0.02 A·m² Explanation: The magnetic moment is the current times the loop’s area: 2 × 0.01 = 0.02 A·m². (20) A particle with charge 2 × 10⁻⁶ C moves at 10 m/s through a 0.5 T field at a 45° angle. What is the magnetic force? a) 7.07 × 10⁻⁶ N b) 1.41 × 10⁻⁵ N c) 2.00 × 10⁻⁵ N d) 2.83 × 10⁻⁵ N  Answer: b) 1.41 × 10⁻⁵ N Explanation: The force depends on charge, speed, field, and the angle’s sine. At 45°, sin(45°) ≈ 0.707. Calculate: 2 × 10⁻⁶ × 10 × 0.5 × 0.707 ≈ 1.41 × 10⁻⁵ N.