A square loop of area 25 cm 2 has a resistance of 10 Ω. The loop is placed in uniform magnetic field of magnitude 40.0 T. The plane of loop is perpendicular to the magnetic field. The work done in pulling the loop out of the magnetic field slowly and uniformly in 1.0 sec, will be

Question

A square loop of area 25 cm 2 has a resistance of 10 Ω. The loop is placed in uniform magnetic field of magnitude 40.0 T. The plane of loop is perpendicular to the magnetic field. The work done in pulling the loop out of the magnetic field slowly and uniformly in 1.0 sec, will be (A) 2.5 × 10-3 J (B) 1.0 × 10-3 J (C) 1.0 × 10-4 J (D) 5 × 10-3 J

Tardigrade Admin · Accepted Answer

Sol. ℓ=50 cm t =1 sec ∴ V =(0.05/1)=0.05 m / s . i =(40 × 0.05 × 0.05/10)=0.01 A ∴ F = B i ℓ=40 × 0.01 × 0.05 F =0.02 N ∴ W =0.02 × ℓ=0.02 × .05 ∴ W =1 × 10-3 J

Q. A square loop of area $25 c m^{2}$ has a resistance of $10Ω$ . The loop is placed in uniform magnetic field of magnitude $40.0 T$ . The plane of loop is perpendicular to the magnetic field. The work done in pulling the loop out of the magnetic field slowly and uniformly in $1.0 sec$ , will be

$2.5 \times 1 0^{- 3} J$

$1.0 \times 1 0^{- 3} J$

$1.0 \times 1 0^{- 4} J$

$5 \times 1 0^{- 3} J$

Sol. $ℓ = 50 c m$
$t = 1 sec$
$∴ V = \frac{0.05}{1} = 0.05 m / s .$
$i = \frac{40 \times 0.05 \times 0.05}{10} = 0.01 A$
$∴ F = B_{i} ℓ = 40 \times 0.01 \times 0.05$
$F = 0.02 N$
$∴ W = 0.02 \times ℓ = 0.02 \times .05$
$∴ W = 1 \times 1 0^{- 3} J$

Q. A square loop of area 25cm2 has a resistance of 10Ω. The loop is placed in uniform magnetic field of magnitude 40.0T. The plane of loop is perpendicular to the magnetic field. The work done in pulling the loop out of the magnetic field slowly and uniformly in 1.0sec, will be

2.5×10−3J

1.0×10−3J

1.0×10−4J

5×10−3J