Question

2. A high-speed electron having a mass of 9 × 10−31𝑘𝑔 is moving at right angle to 0.75-T magnetic field and has a speed of 2.5 × 107𝑚/𝑠. What is the size of the force acts on the high-speed electron? What is the magnitude of acceleration of the particle?

Answer 1

$B=0.75 \;T \\ v=2.5 \times 10^7 \;m/s \\ m = 9 \times 10^{-31} \;kg \\ q = e = 1.6 \times 10^{-19} \;C \\ θ=90°$

Magnetic force

$F = qvBsinθ \\ F = 1.6 \times 10^{-19} \times 2.5 \times 10^7 \times 0.75 \times sin 90° \\ F = 3 \times 10^{-12} \;N$

Acceleration

$a = \frac{F}{m} \\ a = \frac{3 \times 10^{-12}}{9 \times 10^{-31}} \\ a = 0.333 \times 10^{19} = 3.33 \times 10^{18} \;m/s^2$