Find current when linear mass density and current is given.

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SUMMARY

The discussion focuses on calculating the current required in a second wire to balance its weight through magnetic repulsion from a first wire carrying a current of 50 A. The relevant equations include the magnetic field formula, B = μ₀i/(2πr), and the force equation, F = BIL sin(a). The correct current for the second wire, given its linear mass density of 1.0 x 10-4 kg/m and a separation of 5 mm, is determined to be 0.49 A. The calculations involve equating the magnetic force to the gravitational force acting on the second wire.

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arunashish
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Homework Statement


A long straight wire is fixed horizontally and carries a current of 50 A. A second wire having linear mass density 1.0 * 10^{-4} is placed parallel to and directly above this wire at a separation of 5mm. What current should this second wire carry such that the magnetic repulsion can balance its weight?
Equations relevant
(B = μ0i)/2∏r
F = B * I * L * sina
My attempt
I have first got the magnetic field of Wire1 by the formula (B = μ0i)/2∏r.
Then i took out the Mg Field of Wire2 with the same formula taking current as I.Then i took the weight of Wire2 as Total Force and calculated---
Ftotal=B * I * L + B1 * I1 * L1 (taking L as 1m and where B1,L1,I1 as values of Wire2
But i don't get the right answer that is 0.49A.
 
Last edited:
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Hi arunashish. Welcome to PF.

How have you attempted the problem? What do you think would be the equations relevant here?

Please follow the homework-help template while posting questions in this section :smile:
 

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