QJEEAdv17Phb4e9: JEE Advanced 2017 Physics - Class 12 JEEAdv17Phb4e9 Solution: A uniform magnetic field exist in the region between and…

Question

Verified Solution • 2025–26 Session moderate

A uniform magnetic field

B

exist in the region between

x=0

and

x = {{3R} \over 2}

(region

2

in the figure) pointing normally into the plane of the paper. A particle with charge

+Q

and momentum

p

directed along

x

-axis enters region

2

from region

1

at point

{P_1}\left( y \right) = - R).

$ Which of the following option(s) is/are correct?

JEE Advanced 2017 Paper 2 Offline Physics - Magnetism Question 39 English

A For $B > {2 \over 3}{p \over {QR}}$ the particle well re-enter region $1$
B For $B = {8 \over {13}}{p \over {QR}},$ the particle will enter region $3$ through the point ${P_2}$ on $x$ -axis
C When the particle re-enters region $1$ through the longest possible path in region $2,$ the magnitude of the change in its linear momentum between point ${P_1}$ and the farthest point from $y$ -axis is $p/\sqrt 2$
D For a fixed $B,$ particles of same charge $Q$ and same velocity $v,$ the distance between the point ${P_1}$ and the point of re-entry into region $1$ is inversely proportional to the mass of the particle

Accepted Answer

As the magnetic field is perpendicular to motion of particle.

$\therefore$ Particle will move in curved path of radius,

$r = {{mv} \over {QB}} = {p \over {QB}}$ ( $\because$ p = mv)

Particle will cross the region 2 if

$r > {{3R} \over 2} \Rightarrow B < {{2p} \over {3QR}}$

and particle will re-enter region 1 if

$r < {{3R} \over 2} \Rightarrow B > {{2p} \over {3QR}}$

JEE Advanced 2017 Paper 2 Offline Physics - Magnetism Question 39 English Explanation 1

Here a = 0, b = (r $-$ R)

Equation of trajectory, ${(x - a)^2} + {(y - b)^2} = {r^2}$

$\therefore$ To pass through region 2, at ${P_2}\left( {{{3R} \over 2},0} \right)$ ,

${\left( {{{3R} \over 2}} \right)^2} + {(r - R)^2} = {r^2}$ or ${9 \over 4}{R^2} + {r^2} + {R^2} - 2rR = {r^2}$

${{13} \over 4}{R^2} = 2rR$ or $r = {{13} \over 8}R \Rightarrow {p \over {QB}} = {{13} \over 8}R$ or $B = {8 \over {13}}{p \over {QR}}$

Hence, for $B = {8 \over {13}}{p \over {QR}}$ particle will enter region 3 through P₂

For $r < {{3R} \over 2}$ , particle will re-enter region 1 through point P₃(x, y).

$y = (2r - R) = \left( {{{2mv} \over {QB}} - R} \right)$ , x = 0

$\Rightarrow$ y $\propto$ m

JEE Advanced 2017 Paper 2 Offline Physics - Magnetism Question 39 English Explanation 2

At farthest point from y-axis, the momentum is perpendicular to initial momentum.

$\therefore$ $\Delta p = \sqrt 2 mv = \sqrt 2 p$

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