Two long thin parallel wires are placed at a distance (r) from each other carrying equal current  in opposite direction. The magnitude of force per unit length which is exerted by each wire on the other is :

  1. \(\dfrac{\mu_0 I^2}{r}\)
  2. \(\dfrac{\mu_0 I^2}{2\pi r}\)
  3. \(\dfrac{\mu_0 I}{2\pi r}\)
  4. \(\dfrac{\mu_0^2 I}{2\pi r^2}\)

Answer (Detailed Solution Below)

Option 2 : \(\dfrac{\mu_0 I^2}{2\pi r}\)
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NDA 01/2025: English Subject Test
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Detailed Solution

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CONCEPT:

Magnetic field:

  • The space or region around the current-carrying wire/moving electric charge or around the magnetic material in which force of magnetism can be experienced by other magnetic material is called as magnetic field/magnetic induction by that material/current.
  • It is denoted by B.
  • The magnetic force per unit length between two parallel wires is given by;

\(\frac{F}{l} = \frac{{{\mu _o}}}{{2\pi }}\frac{{2{I_1}{I_2}}}{d}\)

Where μ0 = permittivity of free space, I1 = current in a first wire, I2 = current in a second wire,d = distance between two wires, and l = length of current-carrying wire.

EXPLANATION:

​Given - I1 = I2 = I and distance the between the two-wire (d) = r

  • The magnetic force per unit length between two parallel wires is given by;

\(\Rightarrow \frac{F}{l} = \frac{{{\mu _o}}}{{4\pi }}\frac{{2{I_1}{I_2}}}{d}\)

\(\Rightarrow \frac{F}{l} = \frac{{{\mu _o}}}{{4\pi }}\frac{{2{I^2}{}}}{r}=\frac{\mu_o}{2\pi}\frac{I^2}{r}\)

  • As the current in the wire is in the opposite direction, then the force between them will be repulsive.

Important Points

  • If the conductor carries the current in the same direction, then the force between them will be attractive.
  • If the conductor carries the current in the opposite direction, then the force between them will be repulsive
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