Answer to Question #278193 in Electricity and Magnetism for Sya

Question #278193

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.

Expert's answer

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.

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