contestada

At a particular instant, a proton at the origin has velocity < 5e4, -2e4, 0> m/s. You need to calculate the magnetic field at location < 0.03, 0.05, 0 > m, due to the moving proton. What is the vector r?

Respuesta :

Answer:

[tex]9.7\times 10^{-5} T[/tex]

Explanation:

Velocity =[tex]5\times 10^4i-2\times 10^4j[/tex]

r=[tex]0.03i+0.05j[/tex]

r=[tex]\mid r\mid=\sqrt{(0.03)^2+(0.05)^2}=0.058[/tex]

v=[tex]\mid V\mid=\sqrt{(5\times 10^4)^2+(-2\times 10^{4})^2}=5.39\times 10^{2}[/tex]

We know that

[tex]B=\frac{mv}{qr}[/tex]

Where q=[tex]1.6\times 10^{-19} C[/tex]

Mass of proton=[tex]1.67\times 10^{-27} kg[/tex]

Using the formula

[tex]B=\frac{1.67\times 10^{-27}\times 5.39\times 10^2}{1.6\times 10^{-19}\times 0.058}[/tex]

[tex]B=9.7\times 10^{-5} T[/tex]