Magnetic Force on a Current-Carrying Conductor - UNSOLVED PRACTICE SET

Q7. Derive the expression for the magnetic force on a current-carrying conductor: F = I(l × B).

Q8. A wire of length 50 cm carrying current 5 A is placed perpendicular to a magnetic field of 0.4 T. Calculate the force on the wire.

Q9. Two parallel wires carrying currents in opposite directions repel each other. Explain why, using the concept of magnetic fields produced by each wire.

Q10. Why does a current-carrying wire not experience a magnetic force when placed parallel to the magnetic field?

Q11. A straight wire of mass 10 g and length 20 cm is suspended horizontally by two springs. It carries current 2 A in a uniform horizontal magnetic field B = 0.5 T perpendicular to the wire. Calculate the additional extension of each spring if k = 5 N/m.

[Given: g = 9.8 m/s²]

Q12. The force between two parallel current-carrying wires is used to define the ampere. Explain how this definition works.

Q13. Derive the expression for the force between two infinitely long parallel current-carrying conductors. Show that the force is attractive for currents in the same direction and repulsive for opposite directions. Explain how this leads to the definition of one ampere.

Q14. Explain why a current-carrying coil experiences a torque in a magnetic field but a straight wire of the same length experiences a force. Use diagrams to illustrate your answer.

Q15. Two long parallel wires are 10 cm apart in air. Wire 1 carries current I₁ = 5 A upward, and Wire 2 carries I₂ = 3 A upward.

(a) Calculate the force per unit length between the wires. Is it attractive or repulsive?  

(b) Calculate the magnetic field at Wire 2 due to Wire 1.  

(c) If Wire 2 has mass per unit length 0.01 kg/m, what current is needed to support it against gravity using the magnetic force from Wire 1?  

(d) What happens if the direction of I₂ is reversed?

[Given: μ₀ = 4π × 10⁻⁷ T m/A, g = 9.8 m/s²]

Q16. In a school physics lab, a student sets up an experiment to measure the magnetic force on a current-carrying wire. A copper rod of length l = 30 cm and mass m = 50 g rests on two horizontal rails connected to a variable power supply. A uniform magnetic field B = 0.6 T is applied vertically downward. The coefficient of friction between the rod and rails is μ = 0.2.

(a) Draw the free body diagram for the rod when current flows.  

(b) Calculate the minimum current needed to start the rod moving.  

(c) If the current is 5 A, calculate the acceleration of the rod.  

(d) The student places the rod on an inclined plane at 30° to the horizontal, with the magnetic field still vertical. Calculate the current needed to keep the rod stationary.  

(e) Explain why this setup can be used as a simple current-to-force converter.

[Given: g = 9.8 m/s²]

Q17. In an electric train system, two parallel rails carry current to power the train. The rails are separated by d = 1.5 m and each carries I = 500 A in opposite directions.

(a) Calculate the force per unit length between the rails. Is it attractive or repulsive?  

(b) If the rails are 10 km long, calculate the total force between them.  

(c) This force tends to push the rails apart. Explain how railway engineers prevent the rails from spreading.  

(d) If a third rail (return path) is placed midway between the two, carrying the same current, how does the force on each outer rail change?  

(e) Modern high-speed trains use overhead wires instead of third rails. What are the advantages?

[Given: μ₀ = 4π × 10⁻⁷ T m/A]

Q18. A rectangular loop of wire with sides a = 8 cm and b = 12 cm carries current I = 4 A. It is placed in a uniform magnetic field B = 0.5 T.

(a) Calculate the force on each side when the plane of the loop is perpendicular to the field.  

(b) Calculate the net force on the loop.  

(c) Calculate the torque when the plane of the loop makes angle θ = 30° with the field.  

(d) Calculate the maximum torque and the orientation at which it occurs.  

(e) A student claims that the net force on any closed current loop in a uniform magnetic field is always zero. Prove this mathematically for a rectangular loop.