In the $xy$-plane, the region $y>0$ has a uniform magnetic field $B_{1} \widehat{k}$ and the region $y<0$ has another uniform magnetic field $B_{2} \widehat{k}$. A positively charged particle is projected from the origin along the positive $y$-axis with speed $v_{0}=\pi m s^{-1}$ at $t=0$, 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 $B_{2}=4 B_{1}$, the average speed of the particle, in $m s^{-1}$, along the $x$-axis in the time interval $T$ is ____.

JEE Advanced · 2018 · 20 May 2018 · Paper 1 · Magnetic Effects of Current

In the

x y

-plane, the region

y > 0

has a uniform magnetic field

B_{1} k

and the region

y < 0

has another uniform magnetic field

B_{2} k

. A positively charged particle is projected from the origin along the positive

y

-axis with speed

v_{0} = π m s^{- 1}

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} = 4 B_{1}

, the average speed of the particle, in

m s^{- 1}

, along the

x

-axis in the time interval

T

is ____.

Answer: 2

The radius in region 1 is

R = \frac{m v _{0}}{q B _{1}}

and in region 2 is

R /4

. The particle completes a semicircle in each region. Total time:

T = \frac{π m}{q B _{1}} + \frac{π m}{4 q B _{1}} = \frac{5 π m}{4 q B _{1}}

. Total displacement along x:

2 R + R /2 = 5 R /2

. Average speed:

v_{a v g} = \frac{5 R /2}{T} = \frac{5}{2} \cdot \frac{m v _{0}}{q B _{1}} \cdot \frac{4 q B _{1}}{5 π m} = \frac{2 v _{0}}{π} = \frac{2 π}{π} = 2.00

m/s.

JEE Advanced · 2018 · 20 May 2018 · Paper 1 · Numerical answer