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Question:

A uniform magnetic field B exists in the region between x=0 and x=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 the x-axis enters region 2 from region 1 at point P1(y=-R). Which of the following option(s) is/are correct?

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 P1 and the farthest point from y-axis is p/√2

For B=8/(13pQR), the particle will enter region 3 through the point P2 on x-axis

For B > 2p/(3QR), 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 P1 and point of re-entry region 1 is inversely proportional to the mass of the particle

Solution:

The correct options are A, B, and C.

For B=8/(13pQR), the particle will enter region 3 through the point P2 on x-axis
If particle enters regions through P2
Radius = √((3R/2)² + x²) = R + x
9R²/4 + x² = R² + x² + 2Rx
x = 5R/8
Total radius = R + x = 13R/8
Radius is given by p/(qB)
Hence 13R/8 = p/(QB)
B = 8p/(13QR)

Particle re-enters region 1 if radius is less than 3R/2
Hence p/(qB) < 3R/2
B > 2p/(3qR)