Question

A charge of $1.16\times {10}^{-19}$ coulomb enters in the magnetic field of 2 weber/$m^2$ normally with a velocity of 105 m/sec. What is the force on the charge ?

• A. $3.48\times {10}^{-17}$N
• B. $3.21\times {10}^{-19}$N
• C. $1.68\times {10}^{-14}$N
• D. zero

Answer 1

The correct option is A $3.48\times {10}^{-17}$N

The force experiences by a charge particle moving with a velocity $v$ in a magnetic field (B) is: $\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})$
Ignoring the vector notation we can write:
$F=qvB=(1.16\times {10}^{-19}\times 2\times 105)Cweber/ms=3.48\times {10}^{-17}N$