Question

A proton moving with a velocity of $(6i+8j)\times {10}^{5}m{s}^{-1}$ enters uniform magnetic field of induction $5\times {10}^{-3}k$ tesla. The magnitude of the force acting on the proton is $\left(b\right)$ and $k$ are unit vectors forming a right handed traid $)$

• A. Zero
• B. $8\times {10}^{-16}N$
• C. $3\times {10}^{-16}N$
• D. $4\times {10}^{-16}N$

Answer 1

The correct option is B $8\times {10}^{-16}N$

Given: $\overrightarrow{v}=(6i+8j)\ast {10}^{5}m/s$

$\overrightarrow{B}=5\ast {10}^{-3}k$ $T$

To find: magnitude of force,$||\overrightarrow{F}||=?$

Solution: As we know that,

force experienced by a proton in magnetic field is,

$\overrightarrow{F}=e(\overrightarrow{v}\ast \overrightarrow{B})$

$=>\overrightarrow{F}=e\left[\right(6i+8j)\ast (5k)\ast {10}^{5-3}]$

$=>\overrightarrow{F}=1.6\ast {10}^{-19}\ast (4i-3j)\ast {10}^2]$

$=>||\overrightarrow{F}||=\sqrt{(1.6\ast {10}^{-19}{)}^2\ast (4^2+3^2)\ast ({10}^2{)}^2}$

$=>||\overrightarrow{F}||=8\ast {10}^{-17}N$

hence,

The correct option is not present.