Question

A cosmic ray proton moving toward the Earth at 3.80 ✕ 107 m/s experiences a magnetic force of 1.45 ✕ 10−16 N. What is the strength of the magnetic field (in T) if there is a 45° angle between it and the proton's velocity?

Answer 1

`B=3.4* 10^(-29) T`

Step-by-step explanation:

The magnetic force is given by the formula

`F_m=evB\sin\theta`

`e=1.6* 10^-19} ev`, is the proton charge

`\therefore 1.45* 10^(-16)=1.6* 10^(-19)* 3.80* 10^7* B*\sin 45^0`

`\implies 1.45* 10^(-16)=B* 4.2992\tmies 10^(-12)\\\therefore B=(1.45* 10^(-16))/(4.2992\tmies 10^(-12))=3.4* 10^(-29) T`