There is a uniform magnetic field of 1 T. A current carrying loop is kept perpendicular to this field. Find the tension developed in the wire due to the magnetic field. Current in the wire is 1.57 A and the length of the wire is 0.5 m.
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see the diagram enclosed.
We consider a square loop ABCD in a magnetic field B (coming out of the plane of the diagram), carrying a current of i and of length L.
Force on AB of length L/4 = F = i L/4 * B sin 90 = i L B/4
Since the wire is intact, there must be a tension T at points A and B in the wires AD and BC. This will be the same in all the sides of the square loop.
so T = F/2 = i L B/8 = 1.57 * 0.5 * 1 / 8 = 0.098125 N
Tension per unit length = i B/2 = 0.785 N/m
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If the wire is a circular loop then,
R = depends on whether there is one winding or more.
if there is just one winding then: R = L/(2π)
Force on a infinitesimal length dl of wire = i (2RФ) B = 2 T sinФ = 2 T Ф as Ф tends to zero.
so Tension = i R B = i L B / (2 π) = 1.57 * 0.5 * 1 /(2π) = 0.125 N approx.
We consider a square loop ABCD in a magnetic field B (coming out of the plane of the diagram), carrying a current of i and of length L.
Force on AB of length L/4 = F = i L/4 * B sin 90 = i L B/4
Since the wire is intact, there must be a tension T at points A and B in the wires AD and BC. This will be the same in all the sides of the square loop.
so T = F/2 = i L B/8 = 1.57 * 0.5 * 1 / 8 = 0.098125 N
Tension per unit length = i B/2 = 0.785 N/m
============
If the wire is a circular loop then,
R = depends on whether there is one winding or more.
if there is just one winding then: R = L/(2π)
Force on a infinitesimal length dl of wire = i (2RФ) B = 2 T sinФ = 2 T Ф as Ф tends to zero.
so Tension = i R B = i L B / (2 π) = 1.57 * 0.5 * 1 /(2π) = 0.125 N approx.
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