Concepts Of Physics MCQ Edition [Volume 2]PhysicsMagnetic Field due to a Current
The given figure displays a segment of an electric circuit. The long wires AB , CD , and EF possess identical resistances. The separation between adjacent wires is 1.0 cm . Wires AE and BF have negligible resistance, and the ammeter indicates a current of 30 A . Determine the magnetic force per unit length acting on wire AB and wire CD .
Options
- A3 10⁻³ N m ⁻¹ downwards on AB , 2 10⁻³ N m ⁻¹ downwards on CD
- B3 10⁻³ N m ⁻¹ downwards on AB , zero on CD
- C27 10⁻³ N m ⁻¹ downwards on AB , zero on CD
- D2 10⁻³ N m ⁻¹ downwards on AB , zero on CD
Correct answer
B. 3 10⁻³ N m ⁻¹ downwards on AB , zero on CD
Step-by-step solution
Since the wires AB , CD , and EF have identical resistances and are connected in parallel, the total current of 30 A divides equally among them. Current in each wire, I = 30 3 = 10 A . The magnetic force per unit length between two parallel wires carrying currents I₁ and I₂ separated by a distance r is given by f = ₀ I₁ I₂ 2 r . Since the currents are in the same direction, the force is attractive. For wire AB , the forces due to CD and EF are both attractive and act downwards. f_ AB = f_ AB, CD + f_ AB, EF f_ AB =