Test Index

Magnetism

Section: Physics

Question : 18 of 46

Marks: +1, -0

A uniform magnetic field B exists in the region between x = 0 and

x=\frac{3R}{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(y=-R).

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

[JEE Adv 2017 P2]

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

Validate

Solution:

For (A) For

B=\frac{8P}{13QR}

r=\frac{P}{QB}=\frac{p13QR}{Q8p}=\frac{13R}{8}

Clearly

x=\sqrt{\frac{13r^2}{8}-\left(\frac{5R}{8}\right)^2}

x=\frac{3R}{2}

Hence particle enters region 3 through point

P_2

(r = radius of circle in which charge moves)

For (B) The particle will reenter region 1 if

r < \frac{3R}{2}

\frac{P}{QB} < \frac{3R}{2} \Rightarrow B > \frac{2p}{3QR}

For C Distance between point

p_1

and point of re-entry into region 1 = 2r

y=\frac{2mv}{qB} \propto m

For (D) When particle re-enters region 1 through longest possible path, it will re-enter horizontally only.Hence at farthest point from y axis, if must be vertical with magnitude of momentum being p. hence change in its linear momentum must be

\sqrt{2}p

Go to Question:

Prev Question

Next Question