JEE Challenger
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PhyHardAdv 2017 • Paper 2

Charged Particle Trajectory in Uniform Magnetic Field Region

A uniform magnetic field BB is present in the region between x=0x = 0 and x=3R2x = \frac{3R}{2} (region 2 as shown in the figure) directed normally into the plane of the paper. A particle with charge +Q+Q and momentum pp moving along the xx-axis enters region 2 from region 1 at point P1P_1 (y=−Ry = -R). Select the correct statement(s) among the following:

Question Diagram 1

Options

A

For B>23pQRB > \frac{2}{3} \frac{p}{QR}, the charged particle will re-enter region 1

Correct
B

When B=813pQRB = \frac{8}{13} \frac{p}{QR}, the particle enters region 3 through the point P2P_2 located on the xx-axis

Correct
C

If the particle re-enters region 1 after following the longest possible path inside region 2, the magnitude of variation in its linear momentum between point P1P_1 and its farthest distance from the yy-axis equals p/2p/\sqrt{2}

D

For a given value of BB, among particles possessing identical charge QQ and speed vv, the separation distance between P1P_1 and the re-entry location into region 1 varies inversely with the mass of the particle

Topics & Concepts

Moving Charges and MagnetismMagnetic Force

Step-by-Step Solution

To determine the correct option(s), we analyze the motion of the charged particle in the uniform magnetic field.

1. Equation of Motion and Trajectory

  • Magnetic field: B⃗=−Bk^\vec{B} = -B \hat{k} in Region 2 (0≤x≤3R20 \le x \le \frac{3R}{2}).
  • Initial position and momentum: Entering at P1(0,−R)P_1(0, -R) with momentum p⃗=pi^\vec{p} = p\hat{i}, which implies initial velocity v⃗=vi^\vec{v} = v\hat{i}.
  • Magnetic force: F⃗=Q(v⃗×B⃗)=Q(vi^×(−Bk^))=QvBj^\vec{F} = Q(\vec{v} \times \vec{B}) = Q(v\hat{i} \times (-B\hat{k})) = QvB \hat{j}

Since the force acts along the +y+y-direction, the particle undergoes circular motion in the xyxy-plane with radius of curvature: r=mvQB=pQBr = \frac{mv}{QB} = \frac{p}{QB}

Because the velocity at P1(0,−R)P_1(0, -R) is along +i^+\hat{i} and the force is along +j^+\hat{j}, the center of the circular trajectory CC is located at: C=(0,−R+r)C = (0, -R + r)

The equation of the circular trajectory is: x2+(y+R−r)2=r2x^2 + (y + R - r)^2 = r^2

2. Evaluation of Options

Option A:

For the particle to re-enter Region 1 (x≤0x \le 0), the maximum distance it reaches along the xx-axis must be less than the width of Region 2, which is 3R2\frac{3R}{2}.

The maximum xx-coordinate on the circular arc is xmax⁡=rx_{\max} = r. r<3R2  ⟹  pQB<3R2  ⟹  B>23pQRr < \frac{3R}{2} \implies \frac{p}{QB} < \frac{3R}{2} \implies B > \frac{2}{3} \frac{p}{QR}

Thus, Option A is correct.

Option B:

Point P2P_2 is located on the xx-axis (y=0y = 0) at the boundary between Region 2 and Region 3 (x=3R2x = \frac{3R}{2}). Hence, P2=(3R2,0)P_2 = \left(\frac{3R}{2}, 0\right).

Substituting x=3R2x = \frac{3R}{2} and y=0y = 0 into the equation of the trajectory: (3R2)2+(0+R−r)2=r2\left(\frac{3R}{2}\right)^2 + (0 + R - r)^2 = r^2 9R24+R2−2Rr+r2=r2\frac{9R^2}{4} + R^2 - 2Rr + r^2 = r^2 13R24=2Rr  ⟹  r=138R\frac{13R^2}{4} = 2Rr \implies r = \frac{13}{8}R

Using r=pQBr = \frac{p}{QB}: pQB=138R  ⟹  B=813pQR\frac{p}{QB} = \frac{13}{8}R \implies B = \frac{8}{13} \frac{p}{QR}

Thus, Option B is correct.

Option C:

The longest possible path inside Region 2 without escaping into Region 3 corresponds to the maximum possible radius r=3R2r = \frac{3R}{2}.

  • At point P1P_1, the initial momentum is: p⃗1=pi^\vec{p}_1 = p \hat{i}
  • At the farthest point from the yy-axis (x=rx = r), the velocity vector is directed along +j^+\hat{j}, so the momentum is: p⃗farthest=pj^\vec{p}_{\text{farthest}} = p \hat{j}

The magnitude of the change in linear momentum is: ∣Δp⃗∣=∣p⃗farthest−p⃗1∣=∣pj^−pi^∣=p2+p2=2p|\Delta \vec{p}| = |\vec{p}_{\text{farthest}} - \vec{p}_1| = |p\hat{j} - p\hat{i}| = \sqrt{p^2 + p^2} = \sqrt{2}p

Option C states this variation is p/2p/\sqrt{2}, which is incorrect. Thus, Option C is incorrect.

Option D:

To find the point of re-entry into Region 1, we set x=0x = 0 in the trajectory equation: (0)2+(y+R−r)2=r2  ⟹  y+R−r=±r(0)^2 + (y + R - r)^2 = r^2 \implies y + R - r = \pm r

  • y=−Ry = -R (Point P1P_1, entry point)
  • y=2r−Ry = 2r - R (Re-entry point)

The separation distance dd between P1(0,−R)P_1(0, -R) and the re-entry point (0,2r−R)(0, 2r - R) is: d=(2r−R)−(−R)=2r=2mvQBd = (2r - R) - (-R) = 2r = \frac{2mv}{QB}

For given values of BB, QQ, and vv, d∝md \propto m (the distance varies directly with the mass, not inversely). Thus, Option D is incorrect.

Conclusion

The correct options are A and B.

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