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Electromagnetic Induction question

2015 · Shift 0 · Q50
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Electromagnetic Induction question

2015 · Shift 0 · Q50

JEE MainPhysicsElectromagnetic InductionMCQ+4 / −1
Two coaxial solenoids of different radius carry current III in the same direction. F1→\overrightarrow {{F_1}}F1​​ be the magnetic force on the inner solenoid due to the outer one and F2→\overrightarrow {{F_2}}F2​​ be the magnetic force on the outer solenoid due to the inner one. Then :
  1. A
    F1→\overrightarrow {{F_1}}F1​​ is radially in wards and F2→=0\overrightarrow {{F_2}} = 0F2​​=0
  2. B
    F1→\overrightarrow {{F_1}}F1​​ is radially outwards and F2→=0\overrightarrow {{F_2}} = 0F2​​=0
  3. C
    F1→=F2→=0\overrightarrow {{F_1}} = \overrightarrow {{F_2}} = 0F1​​=F2​​=0
  4. D
    F1→\overrightarrow {{F_1}}F1​​ is radially inwards and F2→\overrightarrow {{F_2}}F2​​ is radially outards
View written solutionFree

Correct answer: C

  1. Magnetic field due to an ideal long solenoid

For a long solenoid, the magnetic field is:

B=μ0nIB = \mu_0 n IB=μ0​nI

inside the solenoid, along its axis, and approximately zero outside it.

  1. Force on the inner solenoid due to the outer solenoid

The inner solenoid lies completely inside the outer solenoid. Hence the magnetic field produced by the outer solenoid at the location of the inner one is approximately uniform and parallel to the axis.

For a current element,

dF⃗=I dl⃗×B⃗d\vec F = I\, d\vec l \times \vec BdF=Idl×B

Now in the inner solenoid, the wire is helical. In a uniform axial magnetic field:

  • the azimuthal component of current gives radial forces on different parts,
  • these radial forces cancel over a complete turn,
  • there is no net translational force on the solenoid.

Also, a closed current loop in a uniform magnetic field experiences no net force:

F⃗=I∮dl⃗×B⃗=I(∮dl⃗)×B⃗=0\vec F = I \oint d\vec l \times \vec B = I\left(\oint d\vec l\right)\times \vec B = 0F=I∮dl×B=I(∮dl)×B=0

since

∮dl⃗=0.\oint d\vec l = 0.∮dl=0.

Therefore,

F⃗1=0.\vec F_1 = 0.F1​=0.
  1. Force on the outer solenoid due to the inner solenoid

The field of the inner solenoid outside it is approximately zero (for an ideal long solenoid). The outer solenoid lies mostly outside the inner one, so it experiences essentially no magnetic field due to the inner solenoid.

Hence,

F⃗2=0.\vec F_2 = 0.F2​=0.
  1. Check options
  • A: F⃗1\vec F_1F1​ inward, F⃗2=0\vec F_2=0F2​=0 — incorrect
  • B: F⃗1\vec F_1F1​ outward, F⃗2=0\vec F_2=0F2​=0 — incorrect
  • C: F⃗1=F⃗2=0\vec F_1=\vec F_2=0F1​=F2​=0 — correct
  • D: F⃗1\vec F_1F1​ inward, F⃗2\vec F_2F2​ outward — incorrect
  1. Final answer
F⃗1=F⃗2=0\boxed{\vec F_1 = \vec F_2 = 0}F1​=F2​=0​

So, the correct option is C.

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