An electron is projected into a magnetic field of flux density $10\, Wb/m^2$ with a velocity of $3 × 10^7\, m/s$ at an angle of 30° to the field. The magnetic force on the electron is:

$2.4 × 10^{-11} N$

The correct answer is Option (3) → $2.4 × 10^{-11} N$

Given:

Magnetic flux density, $B = 10\ \text{Wb/m}^2 = 10\ \text{T}$

Velocity of electron, $v = 3 \times 10^7\ \text{m/s}$

Angle between $\vec{v}$ and $\vec{B}$, $\theta = 30^\circ$

Charge of electron, $e = 1.6 \times 10^{-19}\ \text{C}$

Magnetic force is given by:

$F = e\,v\,B\,\sin\theta$

$F = (1.6 \times 10^{-19})(3 \times 10^7)(10)\sin30^\circ$

$F = (1.6 \times 10^{-19})(3 \times 10^8)(\frac{1}{2})$

$F = 2.4 \times 10^{-11}\ \text{N}$

∴ Magnetic force on the electron = $2.4 \times 10^{-11}\ \text{N}$