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:
Answer & explanation
Correct answer: option 3
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 = $2.4 \times 10^{-11}\ \text{N}$