Question:
Force of attraction between two parallel current-carrying conductors is $F$ newton per meter. Current through each of them is doubled and reversed. New force in $N/m$ between these conductors is
Force of attraction between two parallel current-carrying conductors is $F$ newton per meter. Current through each of them is doubled and reversed. New force in $N/m$ between these conductors is
Updated On: Jun 8, 2024
- force of attraction - $4F$
- force of repulsion - $4F$
- force of attraction - $F / 4$
- force of repulsion - $F/4$
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Solution and Explanation
Force per unit length between two parallel wires carrying currents $I_1$ and $I_2$ is given by
$F= \frac{\mu_0 I_1 I_2}{2 \pi r} $
Force is attractive if currents are in same direction. If current through each of them is doubled and reversed, then
$F' = \frac{\mu_0 2 I_1 2 I_2}{2 \pi r} = 4 F$
As direction of current in each conductor is reversed, so again currents in each flow in the same direction. So, force will be attractive.
$F= \frac{\mu_0 I_1 I_2}{2 \pi r} $
Force is attractive if currents are in same direction. If current through each of them is doubled and reversed, then
$F' = \frac{\mu_0 2 I_1 2 I_2}{2 \pi r} = 4 F$
As direction of current in each conductor is reversed, so again currents in each flow in the same direction. So, force will be attractive.
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Concepts Used:
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Magnetism:
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Magnetic Field:
Region in space around a magnet where the Magnet has its Magnetic effect is called the Magnetic field of the Magnet. Let us suppose that there is a point charge q (moving with a velocity v and, located at r at a given time t) in presence of both the electric field E (r) and the magnetic field B (r). The force on an electric charge q due to both of them can be written as,
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