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Question
Two long straight parallel conductors carrying currents I1 and I2 are separated by a distance d. If the currents are flowing in the same direction, show how the magnetic field produced by one exerts an attractive force on the other. Obtain the expression for this force and hence define 1 ampere.
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Solution
Fab = Force experienced by wire 'a' of length 'l' due to the magnetic field of wire 'b'
Fba = Force experienced by wire 'a' of length 'l' due to the magnetic field of wire 'b'
Ba = Magnetic field due to wire 'a'
Bb = Magnetic field due to wire 'b'
Ba = `(mu_0I_a)/(2rd)`
Since `vecF = i(vecl xx vecB)`
`F_(ba) = I_1. l(mu_0I_2)/(2rd)`
Similarly, `F(ab) = I_1.l(mu_0I_2)/(2rd)`
The direction of force experienced by wire 'a' is toward wire 'b'. (As shown in the diagram).
Similarly, the direction of force experienced by wire 'b' is toward wire 'a'. Thus, the force is attractive.
If two long wires are placed in a vacuum at a separation of 1 m, one Ampere would be defined as the current in each wire that would produce a force of 2 × 10–7Nm–1 per unit length of wire.
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