A square loop of side 10 cm and resistance 0.5 Ω is placed vertically in the east-west plane. A uniform magnetic field of 0.10 T is setup across the plane in the north-east direction. The magnetic field is decreased to zero in 0.70 S at a steady rate. Then the magnitude of induced current during th…

Question

A square loop of side 10 cm and resistance 0.5 Ω is placed vertically in the east-west plane. A uniform magnetic field of 0.10 T is setup across the plane in the north-east direction. The magnetic field is decreased to zero in 0.70 S at a steady rate. Then the magnitude of induced current during this time interval will be .

Accepted Answer

(2) 2.0 × 10⁻³ A. Given: side 0.10 m, so A=0.01 m²; R=0.5 Ω; B=0.10 T falling steadily to zero in Δ t=0.70 s (the paper prints the unit as a capital S). Idea: the loop stands in the east-west plane, so its normal points north-south, and a north-east field makes 45° with that normal. Initial flux: Φ=BA cos 45°=0.10×0.01×0.707=7.07×10⁻⁴ Wb; the final flux is zero. Emf: ε=(ΔΦ)/(Δ t)=(7.07×10⁻⁴)/(0.70)=1.01×10⁻³ V. Current: I=ε/R=(1.01×10⁻³)/(0.5). So I≈2.0×10⁻³ A.

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