Question

A cosmic ray proton moving toward the Earth at 6.9 × 10⁷ m/s experiences a magnetic force of 1.3 × 10⁻¹⁶ N. What is the strength of the magnetic field (in uT) if there is a 42° angle between it and the proton’s velocity?

answer to 2 decimal places

Answer 1

Gives

Velocity (v) $=6.9\times10^7m/sec$

Force$(F)=1.3\times10^{-16}N$

$\theta=42°$

We know that

Magnetic force

$F=q(v\times B)$

$F=qvBsin\theta$

$B=\frac{F}{qvsin42°}$

$B=\frac{1.3\times10^{-16}}{1.6\times10^{-19}\times6.9\times10^{7}\times sin42°}$

B=$1.76\times10^{-5}T$

B$=17.6\times10^{-6} T$

B=$17.6\mu T$