Test Index

Magnetism

Section: Physics

© examsnet.com

Question : 16 of 46

Marks: +1, -0

In the xy-plane, the region

y > 0

has a uniform magnetic field

\overset{-}{B_1k}

y < 0

has another uniform magnetic field

\overset{-}{B_2k}

A positively charged particle is projected from the origin along the positive y-axis with speed

v_0 = \pi \text{ m/s}

t = 0

, as shown in the figure. Neglect gravity in this problem. Let

t = T

be the time when the particle crosses the x-axis from below for the first time. If

B_2 = 4B_1

, the average speed of the particle, in

\text{m/s}

, along x - axis in the time interval T is________________.

[JEE Adv 2018 P1]

Your Answer:

Solution:

1)Average speed along x-axis

\langle V_x \rangle = \frac{\int \left|\overset{-}{V}_x\right| \, dt}{\int dt} = \frac{d_1 + d_2}{t_1 + t_2}

2) We have

r_1 = \frac{mv}{qB_1}, r_2 = \frac{mv}{qB_2}

Since

B_1 = \frac{B_2}{4} \therefore r_1 = 4r_2

Time in

B_1 \Rightarrow \frac{\pi m}{qB_1} = t_1

Time in

B_2 \Rightarrow \frac{\pi m}{qB_2} = t_2

Total distance along x-axis

d_1 + d_2 = 2r_1 + 2r_2 = 2(r_1 + r_2) = 2(5r_2)

Total time

T = t_1 + t_2 = 5t_2

∴ Average speed

= \frac{10r_2}{5t_2} = 2 \frac{mv}{qB_2} \times \frac{qB_2}{\pi m} = 2

© examsnet.com

Go to Question:

Prev Question

Next Question