JEE Challenger
More from Moving Charges and Magnetism
PhyMedAdv 2013 • Paper 1

Motion of Charged Particle in Uniform Magnetic Field Region

A particle with mass MM and positive charge QQ, having a constant initial velocity u⃗1=4i^ ms−1\vec{u}_1 = 4\hat{i}\text{ ms}^{-1}, enters a region having a uniform static magnetic field perpendicular to the x-yx\text{-}y plane. The magnetic field region extends from x=0x = 0 to x=Lx = L for all yy. After traversing this region, the particle exits on the opposite side after 10 milliseconds10\text{ milliseconds} with a velocity u⃗2=2(3i^+j^) ms−1\vec{u}_2 = 2(\sqrt{3}\hat{i} + \hat{j})\text{ ms}^{-1}. Which of the following statement(s) is/are correct?

Options

A

The direction of the magnetic field is along the −z-z direction.

Correct
B

The direction of the magnetic field is along the +z+z direction.

C

The magnitude of the magnetic field is 50πM3Q units\dfrac{50\pi M}{3Q}\text{ units}.

Correct
D

The magnitude of the magnetic field is 100πM3Q units\dfrac{100\pi M}{3Q}\text{ units}.

Topics & Concepts

Moving Charges and MagnetismMagnetic Force

Step-by-Step Solution

To determine the correct statements, we analyze the motion of the charged particle in the uniform magnetic field:

1. Direction of the Magnetic Field: The particle has a positive charge Q>0Q > 0 and enters the magnetic field at x=0x = 0 with an initial velocity: u⃗1=4i^ ms−1\vec{u}_1 = 4\hat{i}\text{ ms}^{-1} It exits the region with a velocity: u⃗2=23i^+2j^ ms−1\vec{u}_2 = 2\sqrt{3}\hat{i} + 2\hat{j}\text{ ms}^{-1} Since the yy-component of the velocity increases from 00 to +2 ms−1+2\text{ ms}^{-1}, the magnetic force at the point of entry must be directed along the +y+y direction (i.e., along +j^+\hat{j}).

The magnetic force is given by the Lorentz force formula: F⃗=Q(v⃗×B⃗)\vec{F} = Q(\vec{v} \times \vec{B}) Substituting u⃗1=4i^\vec{u}_1 = 4\hat{i} and B⃗=Bzk^\vec{B} = B_z\hat{k}: F⃗=Q(4i^×Bzk^)=−4QBzj^\vec{F} = Q(4\hat{i} \times B_z\hat{k}) = -4Q B_z\hat{j} For the force to be in the +j^+\hat{j} direction, we must have Bz<0B_z < 0. Thus, the magnetic field is directed along the −z-z direction. Therefore, statement (A) is correct and statement (B) is incorrect.

2. Magnitude of the Magnetic Field: The angle of deflection θ\theta of the particle's velocity vector from its initial direction (along the xx-axis) is given by: tan⁡θ=u2yu2x=223=13  ⟹  θ=π6 rad\tan\theta = \frac{u_{2y}}{u_{2x}} = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}} \implies \theta = \frac{\pi}{6}\text{ rad}

The angular frequency of the circular motion in the magnetic field is: ω=QBM\omega = \frac{QB}{M}

The angle swept by the particle during time t=10 ms=10×10−3 s=1100 st = 10\text{ ms} = 10 \times 10^{-3}\text{ s} = \frac{1}{100}\text{ s} is: θ=ωt=(QBM)t\theta = \omega t = \left(\frac{QB}{M}\right)t

Substituting the known values: π6=(QBM)(1100)\frac{\pi}{6} = \left(\frac{QB}{M}\right) \left(\frac{1}{100}\right)

Solving for the magnetic field magnitude BB: B=100πM6Q=50πM3Q unitsB = \frac{100\pi M}{6Q} = \frac{50\pi M}{3Q}\text{ units}

Therefore, statement (C) is correct and statement (D) is incorrect.

Conclusion: The correct statements are (A) and (C).

PreviousNext