Name: Conversion of a Galvanometer Into a Voltmeter
Uploaded: 2019-07-19
Duration: 7 min 18 s
Description: This video helps us understand the working of a galvanometer. It also teaches us how to convert it into a voltmeter of the desired range both theoretically a...

Question

Derive the expression of force acting per unit length of two long straight parallel current carrying conductors.

Accepted Answer

To derive expression of force per unit length between two long straight parallel conductors: Suppose two long thin straight conductors (or wires) PQ and RS are placed parallel to each other in vacuum (or air) carrying currents $i_{1}$ and $i_{2}$ respectively. It has been observed experimentally that when the currents in the wire are in the same direction, they experience an attractive force and when they carry currents in opposite directions, they experience a repulsive force. Let the conductors PQ and RS carry currents $i_{1}$ and $i_{2}$ in same direction and placed at separation r.

The magnetic field due to current flowing in the first wire, $B_{1} = \frac{μ i_{1}}{2 πr} . . . . . (1)$

According to Maxwell’s right-hand rule or right-hand palm rule, the direction of $B_{1}$ will be perpendicular to the plane of paper and directed downward. Due to this magnetic field, each element of other wire experiences a force. The direction of current element is perpendicular to the magnetic field; therefore the magnetic force on element ab of length $Δ L$ .

$∆ F = B i_{2} ∆ L \sin (90 °) = \frac{μ i_{1} i_{2}}{2 πr} ∆ L$

The Total force on parallel conduct of length $L$ ,

$F = \frac{μ i_{1} i_{2}}{2 πr} L F o r c e p e r u n i t l e n g t h \frac{F}{L} = \frac{μ i_{1} i_{2}}{2 πr}$

Derive the expression of force acting per unit length of two long straight parallel current carrying conductors.

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