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3 Jul 2009 11:44:07 IST

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a horizontal wire of length10 cm and mass 0.3 gm carries a current of 5a . the minimum magnitude of

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a horizontal wire of length10 cm and mass 0.3 gm carries a current of 5a . the minimum magnitude of the magnetic field which can be support the weight of the wire is(g=10m/s^2) give complete solution

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Shrey Choudhary

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4 Jul 2009 00:45:19 IST

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length = 10cm = 0.1m = l (small L)

mass = 0.3gm = 3 x 10 ^ -4 kg = m

Weight of the wire (wt) = m x g = 3 x 10 ^ -3 = Minimum Force applied by the magnetic field. (F)

F=Wt

=> I x L x B = wt

=> B _min=wt / I x L

B min = 0.6 x 10 ^ -2

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Sushobhan Sen

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15 Jul 2009 00:22:56 IST

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a horizontal wire of length10 cm and mass 0.3 gm carries a current of 5a . the minimum mag

Downward force = mg = 3 x 10^-4 x 10 = 3 x 10^-3 N

Upward force = BIL = B x 5 x 10x10^-2 = B/2

Equating the two at equilibrium, B = 6 x 10^-3 T

(This is not the complete answer. For the complete answer, mention the direction of the current)

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