Answer to Question #196031 in Electricity and Magnetism for Kenma

Question

A proton moves at 4.45 × 10⁷ m/s perpendicular to a magnetic field. The field causes the proton to travel in a circular path of radius of 0.09 m. What is the field strength? (m_proton = 1.6726 × 10^-27 kgs and q=1.602×10⁻¹⁹ C)

Answer 1

The proton experiences the action of Lorentz force:

$F = qvB$

At the same time, according to the second Newton's law, for a particle moving on the circular orbit

$\displaystyle F = ma = \frac{mv^2}{R}$

$\displaystyle \frac{mv^2}{R} = qvB$

$\displaystyle B=\frac{mv}{qR}= \frac{1.6726 \cdot 10^{-27} \cdot 4.45 \cdot 10^7}{1.602 \cdot 10^{-19} \cdot 0.09} =5.1623 \; [T]$

Answer: 5.1623 T