JEE MAIN - Physics (2020 - 3rd September Morning Slot - No. 20)

A charged particle carrying charge 1 $$\mu $$C is moving
with velocity $$\left( {2\widehat i + 3\widehat j + 4\widehat k} \right)$$ ms–1. If an external
magnetic field of $$\left( {5\widehat i + 3\widehat j - 6\widehat k} \right)$$× 10–3 T exists in the region where the particle is moving then the
force on the particle is $$\overrightarrow F $$ × 10–9 N. The vector $$\overrightarrow F $$ is :
$${ - 0.30\widehat i + 0.32\widehat j - 0.09\widehat k}$$
$${ - 300\widehat i + 320\widehat j - 90\widehat k}$$
$${ - 30\widehat i + 32\widehat j - 9\widehat k}$$
$${ - 3.0\widehat i + 3.2\widehat j - 0.9\widehat k}$$

Explanation

Given,

$${\overrightarrow V }$$ = $$\left( {2\widehat i + 3\widehat j + 4\widehat k} \right)$$ ms–1

$${\overrightarrow B }$$ = $$\left( {5\widehat i + 3\widehat j - 6\widehat k} \right)$$× 10–3 T

q = 1 $$\mu $$C

$$\overrightarrow F = q\left( {\overrightarrow V \times \overrightarrow B } \right)$$

= $${10^{ - 6}} \times {10^{ - 3}} \times \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 2 & 3 & 4 \cr 5 & 3 & { - 6} \cr } } \right|$$

= ($${ - 30\widehat i + 32\widehat j - 9\widehat k}$$) $$ \times $$ 10-9

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