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?

Respuesta :

Answer:

[tex]B=3.4\times 10^{-29} T[/tex]

Explanation:

The magnetic force is given by the formula

[tex]F_m=evB\sin\theta[/tex]

[tex]e=1.6\times 10^-19} ev[/tex], is the proton charge

[tex]\therefore 1.45\times 10^{-16}=1.6\times 10^{-19}\times 3.80\times 10^7\times B\times\sin 45^0[/tex]

[tex]\implies 1.45\times 10^{-16}=B\times 4.2992\tmies 10^{-12}\\\therefore B=\frac{1.45\times 10^{-16}}{4.2992\tmies 10^{-12}}=3.4\times 10^{-29} T[/tex]