JEE Main · 2020 · 03 Sep 2020 · Shift 1 · Magnetic Effects of Current

A charged particle carrying charge

1 μ

C is moving with velocity

(2 \hat{i} + 3 \hat{j} + 4 \hat{k})

^{- 1}

. If an external magnetic field of

(5 \hat{i} + 3 \hat{j} - 6 \hat{k}) \times 1 0^{- 3}

T exists in the region where the particle is moving then the force on the particle is

F \times 1 0^{- 9}

N. The vector

F

is:

Answer: (A)

Solution

The magnetic force is

F = q v \times B

. The cross product

(2 \hat{i} + 3 \hat{j} + 4 \hat{k}) \times (5 \hat{i} + 3 \hat{j} - 6 \hat{k})

has components

\hat{i} [3 (- 6) - 4 (3)] = - 30

\hat{j} [- (2 (- 6) - 4 (5))] = 32

and

\hat{k} [2 (3) - 3 (5)] = - 9

, i.e.

(- 30 \hat{i} + 32 \hat{j} - 9 \hat{k})

. Multiplying by

q = 1 0^{- 6}

C and the field factor

1 0^{- 3}

gives

F = 1 0^{- 9} (- 30 \hat{i} + 32 \hat{j} - 9 \hat{k})

N, so the vector written as

F \times 1 0^{- 9}

- 30 \hat{i} + 32 \hat{j} - 9 \hat{k}

JEE Main · 2020 · 03 Sep 2020 · Shift 1 · Single correct option