Question

A rectangular loop of sides 20 cm and 10 cm carries a current of 5.0 A. A uniform magnetic field of magnitude 0.20 T exists parallel to the longer side of the loop. (a) What is the force acting on the loop? (b) What is the torque acting on the loop?

Answer 1

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Answer:

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l = 20cm = 20 × 10–2m B = 10cm = 10 × 10–2m i = 5A, B = 0.2T (a) There is no force on the sides AB and CD. But the force on the sides AD and BC are opposite. So they cancel each other. (b) Torque on the loop τ = ni vector A x vector B = niABSin90° = 1 × 5 × 20 × 10–2 × 10× 10–2 0.2 = 2 × 10–2 = 0.02N-M Parallel to the shorter side.Read more on Sarthaks.com - https://www.sarthaks.com/66376/a-rectangular-loop-of-sides-20cm-and-10cm-carries-a-current-of-5-0a

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