An electron is moving along +x direction with a velocity of 6 \times 106 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 :

5 \times 10–3 T, along +z direction

3 \times 10–4 T, along –z direction

5 \times 10–3 T, along –z direction

3 \times 10–4 T, along +z direction

Magnetics PYQ — JEE Main 2020 · 6 Sep · Shift 1 · Q49 with Solution

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Correct answer: C

For the electron to keep moving along the $+x$ direction, the net force on it must be zero:

\vec F = q(\vec E + \vec v \times \vec B)=0

Since $q \neq 0$ , we need

\vec E + \vec v \times \vec B = 0

So,

\vec v \times \vec B = -\vec E

Velocity of electron: $\vec v = 6 \times 10^6\, \hat i\; \text{m s}^{-1}$
Electric field: $E = 300\, \text{V/cm} = 300 \times 100 = 3 \times 10^4\, \text{V/m}$ and direction is $+y$ , so $\vec E = 3 \times 10^4\, \hat j\; \text{V/m}$

We need

\vec v \times \vec B = -\vec E

Since $\vec E$ is along $+\hat j$ , we require

\vec v \times \vec B \text{ along } -\hat j

Now $\vec v$ is along $+\hat i$ .

Using cross product:

\hat i \times \hat k = -\hat j

So $\vec B$ must be along $+\hat k$ , i.e. $+z$ direction.

For zero net force,

E = vB

Therefore,

B = \frac{E}{v} = \frac{3 \times 10^4}{6 \times 10^6}

B = 0.5 \times 10^{-2} = 5 \times 10^{-3}\, \text{T}

Thus the magnetic field should be:

5 \times 10^{-3}\, \text{T, along } +z\text{ direction}

So the correct option is C.

Stored correct answer: C

My derived answer: C

They agree.