An electron, moving along the $x$-axis with an initial energy of 100 eV , enters a region of magnetic field $\vec{B}=\left(1.5 \times 10^{-3} T\right) \hat{k}$ at $S$ (See figure). The field extends between $x=0$ and $x=2 \mathrm{~cm}$. The electron is detected at the point $Q$ on a screen placed 8 cm away from the point S . The distance d between P and Q (on the screen) is : (electron's charge $=1.6 \times 10^{-19} \mathrm{C}$, mass of electron $=9.1 \times 10^{-31} \mathrm{kg}$)

JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Magnetic Effects of Current

An electron, moving along the

x

-axis with an initial energy of 100 eV , enters a region of magnetic field

B = (1.5 \times 1 0^{- 3} T) \hat{k}

S

(See figure). The field extends between

x = 0

and

x = 2 cm

. The electron is detected at the point

Q

on a screen placed 8 cm away from the point S . The distance d between P and Q (on the screen) is : (electron's charge

= 1.6 \times 1 0^{- 19} C

, mass of electron

= 9.1 \times 1 0^{- 31} kg

)

Answer: (A)

The electron's speed comes from its kinetic energy, so inside the field it moves on a circle of radius

r = \frac{2 m K}{e B} = \frac{2 ( 9.1 \times 1 0 ^{- 31} ) ( 100 \times 1.6 \times 1 0 ^{- 19} )}{( 1.6 \times 1 0 ^{- 19} ) ( 1.5 \times 1 0 ^{- 3} )} \approx 2.25

cm. Crossing the

2

cm wide field strip it turns through an angle

θ

with

sin θ = \frac{2}{2.25}

, so the sideways shift at the exit of the strip is

r (1 - cos θ) \approx 1.22

cm. Past the field the path is straight, inclined at

θ

to the axis; over the remaining

8 - 2 = 6

cm up to the screen this contributes a further

6 tan θ \approx 11.65

cm. Point

P

is where the undeflected beam would land and

Q

where the electron actually lands, so their separation is the sum of the in-field shift and this straight-line projection,

1.22 + 11.65 \approx 12.87

cm.

JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Single correct option