A uniform magnetic field B exists in a direction perpendicular to the plane of a square loop made of a metal wire. The wire has a diameter of 4 mm and a total length of 30 cm. The magnetic field changes with time at a steady rate dB/dt = 0.032 Ts –1 . The induced current in the loop is close to (Resistivity of the metal wire is 1.23 × 10 –8 Ωm)

Solution

Physics sir · 3 steps

The same 30 cm of wire sets both the loop's area for the emf and the wire's resistance, so use it twice.

1. Induced emf. The 30 cm wire is bent into a square, so each side is 0.3/4 = 0.075 m. The loop area is (0.075)² = 5.625 × 10⁻³ m², so the emf is ε = AdB/dt = 5.625 × 10⁻³ × 0.032 = 1.8 × 10⁻⁴ V.
2. Wire resistance. The radius is half the diameter: r = 2 mm = 2 × 10⁻³ m. The cross-section is π r² = π × 4 × 10⁻⁶ = 1.257 × 10⁻⁵ m², so R = (ρ L)/(π r²) = (1.23 × 10⁻⁸ × 0.3)/(1.257 × 10⁻⁵) = 2.94 × 10⁻⁴ Ω.

   I = ε/R = (1.8 × 10⁻⁴ V)/(2.94 × 10⁻⁴ Ω) ≈ 0.61 A

3. So. The induced current is close to 0.61 A, which is option (d).

The idea to keep. For a loop of wire in a changing field, find the emf from the loop's area and the resistance from the wire's length and cross-section. Then I = ε/R. Always halve the diameter to get the radius.

Answer from the NTA final answer key. Solution by PrepWiser.