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Concepts Of Physics MCQ Edition [Volume 2]
Exam
Concepts Of Physics MCQ Edition [Volume 2]
Subject
Physics
Chapter
Magnetic Field due to a Current

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Magnetic Field due to a Current

A long, tightly-wound solenoid of radius r has n turns per unit...

A long, tightly-wound solenoid of radius rr has nn turns per unit length and carries a current ii. A particle of mass mm and charge qq is projected from a point on the axis in a direction perpendicular to the axis. What is the maximum speed the particle can have such that it does not strike the solenoid?

Options

  1. Aμ0qrnim\dfrac{\mu_0 q r n i}{m}
  2. Bμ0qrni4m\dfrac{\mu_0 q r n i}{4m}
  3. C2μ0qrnim\dfrac{2\mu_0 q r n i}{m}
  4. Dμ0qrni2m\dfrac{\mu_0 q r n i}{2m}

Correct answer

D. μ0qrni2m\dfrac{\mu_0 q r n i}{2m}

Step-by-step solution

The magnetic field inside a long solenoid is uniform and directed along its axis.

B=μ0niB = \mu_0 n i

When a particle of charge qq and mass mm is projected from the axis perpendicular to the magnetic field with speed vv, it moves in a circular path. The radius of this circular path is given by:

R=mvqBR = \dfrac{mv}{qB}

Since the particle is projected from the axis, the axis lies on the circumference of its circular trajectory. The maximum distance of the particle from the axis is equal to the diameter of its circular path, which is 2R2R.

For the particle to not strike the solenoid, this maximum distance must be less than or equal to the radius of the solenoid rr.

2R≤r2R \le r

2(mvqB)≤r2 \left( \dfrac{mv}{qB} \right) \le r

v≤qBr2mv \le \dfrac{qBr}{2m}

Substituting the value of the magnetic field B=μ0niB = \mu_0 n i:

v_\max = \dfrac{\mu_0 q r n i}{2m}
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