A square loop of area 25 cm 2 has a resistance of 10Ω . This loop is placed in a uniform magnetic field of magnitude 40 T . The plane of loop is perpendicular to the magnetic field. The work done in p

Question

A square loop of area 25 cm 2 has a resistance of 10Ω . This loop is placed in a uniform magnetic field of magnitude 40 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 one second, will be

TestHub · Accepted Answer

Given: area of square loop = 25 cm ∴ l = 25 ​ = 5 cm ⇒ 0.05 m ∴ ∴ ∴ ​ R = 10Ω , t = 1 sec , B = 40 T Velocity v = t l ​ = 1 0.05 ​ = 0.05 m / s Motional emf ε m a x ​ = B / v I = R ε ​ = R B / v ​ I = 10 40 × 0.05 × 0.05 ​ = 0.01 A ​ We know, Force acting on loop ∴ F = BI l Using W = F . s , Work done W ​ = 40 × 0.01 × 0.05 = 0.02 N = BI l × l = 0.02 × 0.05 = 1 × 1 0 − 3 J ​

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