An electron is moving along $+x$ direction with a velocity of $6 \times 10^{6} \mathrm{~ms}^{-1}$. It enters a region of uniform electric field of $300 \mathrm{~V} / \mathrm{cm}$ pointing along +y direction. The magnitude and direction of the magnetic field set up in this region such that the electron keeps moving along the $x$ direction will be:

(C) $5 \times 10^{-3} \mathrm{~T}$, along +z direction

(A) $3 \times 10^{-4} \mathrm{~T}$, along +z direction

(B) $5 \times 10^{-3} \mathrm{~T}$, along -z direction

(D) $3 \times 10^{-4} \mathrm{~T}$, along -z direction

JEE Main · 2020 · 06 Sep 2020 · Shift 1 · Magnetic Effects of Current

An electron is moving along

+ x

direction with a velocity of

6 \times 1 0^{6} ms^{- 1}

. It enters a region of uniform electric field of

300 V / cm

pointing along +y direction. The magnitude and direction of the magnetic field set up in this region such that the electron keeps moving along the

x

direction will be:

Answer: (C)

For the electron to keep moving undeflected along

x

, the electric and magnetic forces on it must cancel, which needs

E + v \times B = 0

, i.e.

v \times B = - E

. The field is

E = 300 V/cm = 3 \times 1 0^{4} \hat{j} V/m

and the velocity is

v = 6 \times 1 0^{6} \hat{i} m/s

. Taking

B = B \hat{k}

gives

v \times B = v B (\hat{i} \times \hat{k}) = - v B \hat{j}

, which points along

- \hat{j}

as required, so

B

lies along

+ z

. Equating magnitudes,

v B = E

, hence

B = \frac{E}{v} = \frac{3 \times 1 0 ^{4}}{6 \times 1 0 ^{6}} = 5 \times 1 0^{- 3} T

directed along

+ z

JEE Main · 2020 · 06 Sep 2020 · Shift 1 · Single correct option