Answer to Question #301477 in Electricity and Magnetism for aisme

Question

An electron moves in a uniform magnetic field. At a particular instant the speed of the

electron is 3.0x10^(6) m/s and the magnitude of the magnetic force acting on it is

5.0x10^(-13) N. The velocity of the electron makes an angle of 30^o with the direction of

the magnetic field. Calculate the magnetic field strength at that instant.

Answer 1

Answer

the magnetic field strength at that instant

$B=\frac{F}{qvcos\theta}\\=\frac{5*10^{-13}}{1.6*10^{-19}*3*10^6*cos30°}\\=1.21T$