Answer to Question #278193 in Electricity and Magnetism for Sya

Question

A proton moving with a velocity of 3×10⁸ ms‐¹ perpendicularly in a magnetic field strength X T. Given that the force of the proton received is 4.8×10‐¹³ N. Find the value of X.

Answer 1

We can find the magnetic field strength as follows:

$$F=qvBsin\theta,$$ $$B=\dfrac{F}{qvsin\theta}=\dfrac{4.8\times10^{-13}\ N}{1.6\times10^{-19}\ C\times3\times10^8\ \dfrac{m}{s}\times sin90^{\circ}},$$ $$B=0.01\ T.$$