Question

a) An electron moving at 5 x 10³ m/s in a 1.5T magnetic field experiences a magnetic force of 1.6× 10⁻¹⁶N. What angle does the velocity of the electron make with the magnetic field? 

b) A cosmic ray proton moving toward the earth at 5 x 10⁷ m/s experiences a magnetic force of 1.70 × 10 ⁻¹⁶N. What is the strength of the magnetic field if there is a 45° angle between it and the proton’s velocity? 

Answer 1

Part(a)

We know that

$F=qvBsin\theta$

$\theta=sin^{-1}(\frac{F}{qvB})$

$\theta=sin^{-1}(\frac{1.6\times10^{-16}}{-1.6\times10^{-19}\times 5\times10^{3}\times1.5})$

$\theta=sin^{-1}(-0.13)=-7.66°$

Part (b)

$F=qvBsin\theta$

$B=\frac{F}{qvsin\theta}$

$$B=\frac{1.7\times10^{-16}}{1.6\times10^{-19}\times5\times10^{7}\times sin45°}=3\times10^{-5}T$$