Questions
A loop of flexible conducting wire of length l lies in magnetic field B which is normal to the plane of loop. A current i is passed through the loop. The tension developed in the wire to open up is
detailed solution
Correct option is C
Bi(dl)=2Tsin(dθ/2) ⇒ Bi(rdθ)=2T(dθ/2) (θ is small, sinθ=θ) ⇒ T=Bir=Bil/2πTalk to our academic expert!
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Figure shows a conducting loop ABCDA placed in a uniform magnetic field (strength B) perpendicular to its plane. The part ABC is the (three-fourth) portion of the square of side length l. The part ADC is a circular arc of radius R. The points A and C are connected to a, battery which supplies a current i to the circuit.The magnetic force on the loop due to the field B is
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