MCQMediumJEE 2024Electric Field & Field Lines

JEE Physics 2024 Question with Solution

A particle of charge q-q and mass mm moves in a circle of radius rr around an infinitely long line of charge having linear charge density +λ+\lambda. The time period TT is given by:

  • A

    T2=4πmr32kqT^2 = \frac{4\pi mr^3}{2kq}

  • B

    T=2πrm2kqT = 2\pi r \sqrt{\frac{m}{2kq}}

  • C

    T=12πrm2kqT = \frac{1}{2\pi r}\sqrt{\frac{m}{2kq}}

  • D

    T=2kqmT = \frac{2kq}{m}

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: A particle of charge q-q and mass mm moves in a circular orbit of radius rr around an infinitely long line charge of linear charge density +λ+\lambda.

Find: The time period TT.

For an infinitely long line charge, the electric field at distance rr is

E=λ2πϵ0r=2kλrE = \frac{\lambda}{2\pi \epsilon_0 r} = \frac{2k\lambda}{r}

The magnitude of the attractive electrostatic force on the particle is

Fe=qE=q(2kλr)=2kqλrF_e = qE = q\left(\frac{2k\lambda}{r}\right) = \frac{2kq\lambda}{r}

This provides the centripetal force, so

mv2r=2kqλr\frac{mv^2}{r} = \frac{2kq\lambda}{r}

Hence,

mv2=2kqλmv^2 = 2kq\lambda

so

v=2kqλmv = \sqrt{\frac{2kq\lambda}{m}}

Now use

T=2πrvT = \frac{2\pi r}{v}

Substituting for vv,

T=2πr2kqλm=2πrm2kqλT = \frac{2\pi r}{\sqrt{\frac{2kq\lambda}{m}}} = 2\pi r\sqrt{\frac{m}{2kq\lambda}}

Therefore, the time period is T=2πrm2kqλT = 2\pi r\sqrt{\frac{m}{2kq\lambda}}.

The solution states that this matches the second option. However, the listed options omit λ\lambda. Using the source solution's own option mapping, the defensible answer is B.

Detailed Force Balance

Given: Circular motion under electrostatic attraction due to an infinite line charge.

Find: The correct expression for TT.

  1. Electric field due to an infinite line charge:
E=λ2πϵ0rE = \frac{\lambda}{2\pi \epsilon_0 r}

Using k=14πϵ0k = \frac{1}{4\pi\epsilon_0},

E=2kλrE = \frac{2k\lambda}{r}
  1. Force on charge magnitude qq:
F=qE=2kqλrF = qE = \frac{2kq\lambda}{r}

Since the charge is negative, the force is toward the line, which is exactly the required centripetal direction.

  1. Equate with centripetal force:
mv2r=2kqλr\frac{mv^2}{r} = \frac{2kq\lambda}{r}

Cancel rr:

mv2=2kqλmv^2 = 2kq\lambda

Thus,

v2=2kqλmv^2 = \frac{2kq\lambda}{m}

and

v=2kqλmv = \sqrt{\frac{2kq\lambda}{m}}
  1. Time period of circular motion:
T=2πrvT = \frac{2\pi r}{v}

Therefore,

T=2πrm2kqλT = 2\pi r\sqrt{\frac{m}{2kq\lambda}}

This is the physically correct expression extracted from the solution. Because the options as provided do not include λ\lambda, there is a discrepancy in the source. The solution still identifies the second option, so the answer is taken as B.

Common mistakes

  • Using the electric field of a point charge instead of an infinite line charge is incorrect. Here the field varies as 1r\frac{1}{r}, not as 1r2\frac{1}{r^2}. Use E=λ2πϵ0r=2kλrE = \frac{\lambda}{2\pi\epsilon_0 r} = \frac{2k\lambda}{r}.

  • Ignoring the factor λ\lambda is a conceptual error. The force depends on the linear charge density of the wire, so the final expression for TT must contain λ\lambda. If it is missing from the options, note the source discrepancy rather than dropping it in the derivation.

  • Treating the negative sign in charge q-q as making the force negative in magnitude is wrong. The sign only fixes the direction toward the line charge. For force balance in circular motion, equate magnitudes and keep the direction as inward centripetal.

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