Question

A point charge of 10C moves with a uniform velocity of $2\widehat{a_x}-4\widehat{a_z}m/s$ in electromagnetic field having $\overrightarrow{E}=\widehat{a_x}-3\widehat{a_y}+4\widehat{a_z}V/mand\hat{B}=0.3\widehat{a_x}+0.1\widehat{a_y}Wb/m^2$. The magnitude of force exerted on charge is

• A. 61.02 mN
• B. 159.70 mN
• C. 61.02 N
• D. 159.70 N

Answer 1

The correct option is C 61.02 N

Net force of charge is given as

$\overrightarrow{F}={\overrightarrow{F}}_e+{\overrightarrow{F}}_m$

$\overrightarrow{F}=q\overrightarrow{E}+q(\overrightarrow{\upsilon }\times \overrightarrow{B})N$

$\overrightarrow{F}=10[{\hat{a}}_x-3{\hat{a}}_y+4{\hat{a}}_z+(2{\hat{a}}_x-4{\hat{a}}_z)\times (0.3{\hat{a}}_x+0.1{\hat{a}}_y\left)\right]$

$=10[{\hat{a}}_x-3{\hat{a}}_y+4{\hat{a}}_z+0.2{\hat{a}}_z-1.2{\hat{a}}_y+0.4{\hat{a}}_x]$

$=10[1.4{\hat{a}}_x-4.2{\hat{a}}_y+4.2{\hat{a}}_z]$

$\overrightarrow{F}=14{\hat{a}}_x-42{\hat{a}}_y+42{\hat{a}}_z$

$\left|\overrightarrow{F}\right|=\sqrt{(14{)}^2+(42{)}^2+(42{)}^2}=61.02N$