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Magnetics question

2005 · Shift 0 · Q164
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Magnetics question

2005 · Shift 0 · Q164

JEE MainPhysicsMagneticsMCQ+4 / −1
Two thin, long, parallel wires, separated by a distance ′d′'d'′d′ carry a current of ′i′A'i'A′i′A in the same direction. They will
  1. A
    repel each other with a force of μ0i2/(2πd){\mu _0}{i^2}/\left( {2\pi d} \right)μ0​i2/(2πd)
  2. B
    attract each other with a force of μ0i2/(2πd){\mu _0}{i^2}/\left( {2\pi d} \right)μ0​i2/(2πd)
  3. C
    repel each other with a force 0i2/(2πd2)_0{i^2}/\left( {2\pi {d^2}} \right)0​i2/(2πd2)
  4. D
    attract each other with a force of μ0i2/(2πd2){\mu _0}{i^2}/\left( {2\pi {d^2}} \right)μ0​i2/(2πd2)
View written solutionFree

Correct answer: B

  1. Magnetic field produced by one long straight wire

For a long straight wire carrying current iii, the magnetic field at a perpendicular distance ddd is

B=μ0i2πdB = \frac{\mu_0 i}{2\pi d}B=2πdμ0​i​
  1. Force on the second wire

If the second parallel wire also carries current iii, then the magnetic force on a length LLL of this wire is

F=iLBF = iLBF=iLB

Substituting BBB:

F=iL⋅μ0i2πd=μ0i2L2πdF = iL \cdot \frac{\mu_0 i}{2\pi d} = \frac{\mu_0 i^2 L}{2\pi d}F=iL⋅2πdμ0​i​=2πdμ0​i2L​

Hence, the force per unit length is

FL=μ0i22πd\frac{F}{L} = \frac{\mu_0 i^2}{2\pi d}LF​=2πdμ0​i2​
  1. Nature of force

For two parallel wires carrying current in the same direction, the force between them is attractive.

So the wires attract each other with force per unit length

μ0i22πd\frac{\mu_0 i^2}{2\pi d}2πdμ0​i2​
  1. Checking options
  • A: Repel, with μ0i22πd\dfrac{\mu_0 i^2}{2\pi d}2πdμ0​i2​ — incorrect nature
  • B: Attract, with μ0i22πd\dfrac{\mu_0 i^2}{2\pi d}2πdμ0​i2​ — correct
  • C: Repel, with μ0i22πd2\dfrac{\mu_0 i^2}{2\pi d^2}2πd2μ0​i2​ — incorrect magnitude and nature
  • D: Attract, with μ0i22πd2\dfrac{\mu_0 i^2}{2\pi d^2}2πd2μ0​i2​ — incorrect magnitude

Therefore, the correct answer is B.

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