Question

Force experienced by the charged particle in units of 10^-9 N will be ?

Answer 1

Answer:

$\boxed{\mathfrak{(2) \ - 30\hat{i} + 21 \hat{j} - 2 \hat{k}}}$

Given:

$\sf Charge \: (q)= 1 \mu C = 1 \times {10}^{ - 6} \: C$

$\sf Velocity ( \overrightarrow{v}) = 4 \hat{i} + 6 \hat{j} + 3 \hat{k}$

$\sf Uniform \: magnetic \: field \: ( \overrightarrow{B}) = 3\hat{i} + 4 \hat{j} - 3 \hat{k} \times {10}^{ - 3}$

To Find:

Force ($\sf \overrightarrow{F}$) experience by charged particle in units of $\sf 10^{-9}$ N

Explanation:

Force exerted by a magnetic field is called magnetic force. It is given by:

$\boxed{ \bold{\overrightarrow{F} = q(\overrightarrow{v} \times \overrightarrow{B})}}$

By substituting value of q, $\sf \overrightarrow{v}$ & $\sf \overrightarrow{B}$ in the equation we get:

$\sf \implies \overrightarrow{F} = 1 \times {10}^{ - 6} ((4 \hat{i} + 6 \hat{j} + 3 \hat{k}) \times (3\hat{i} + 4 \hat{j} - 3 \hat{k} \times {10}^{ - 3} )) \\ \\ \sf \implies \overrightarrow{F} = {10}^{ - 9} ((4 \hat{i} + 6 \hat{j} + 3 \hat{k}) \times (3\hat{i} + 4 \hat{j} - 3 \hat{k})) \\ \\ \sf \implies \overrightarrow{F} = (( 6 \times ( - 3) - 3 \times 4) \hat{i} - (4 \times ( - 3) - 3 \times 3) \hat{j} + (4 \times 4 - 6 \times 3)\hat{k}) \times {10}^{ - 9} \\ \\ \sf \implies \overrightarrow{F} = (( - 18 - 12)\hat{i} - ( - 12 - 9) \hat{j} + (16 - 18) \hat{k}) \times {10}^{ - 9} \\ \\ \sf \implies \overrightarrow{F} = (- 30 \hat{i} - ( - 21) \hat{j} + ( - 2) \hat{k}) \times {10}^{ - 9} \\ \\ \sf \implies \overrightarrow{F} = (- 30\hat{i} + 21 \hat{j} - 2 \hat{k}) \times {10}^{ - 9} \: N$

$\therefore$

Force ($\sf \overrightarrow{F}$) experience by charged particle in units of $\sf 10^{-9}$ N = $\sf - 30\hat{i} + 21 \hat{j} - 2 \hat{k}$

Answer 2

Answer:-

$\bf {\overrightarrow{ F} = -30\hat{ i}+21\hat {j}-2\hat{K} \times {10}^{ - 9} N}$

Refer the attachment!!!