Question

An electron is accelerated by a 5.9 kV potential difference. das (sd38882) – Homework #9 – yu – (44120) 3 The charge on an electron is 1.60218 × 10−19 C and its mass is 9.10939 × 10−31 kg. How strong a magnetic field must be experienced by the electron if its path is a circle of

Answer 1

Complete Question

An electron is accelerated by a 5.9 kV potential difference. das (sd38882) – Homework #9 – yu – (44120) 3 The charge on an electron is 1.60218 × 10−19 C and its mass is 9.10939 × 10−31 kg. How strong a magnetic field must be experienced by the electron if its path is a circle of radius 5.4 cm?

Answer:

The magnetic field strength is  $B= 0.0048 T$

Explanation:

The work done by the potential difference on the electron is related to the kinetic energy of the electron by this mathematical expression

                             $\Delta V q = \frac{1}{2}mv^2$

      Making v the subject

                             $v = \sqrt{[\frac{2 \Delta V * q }{m}] }$

 Where m is the mass of electron

              v is the velocity of electron

              q charge on electron

               $\Delta V$ is the potential difference  

Substituting values

         $v = \sqrt{\frac{2 * 5.9 *10^3 * 1.60218*10^{-19} }{9.10939 *10^{-31]} }$f

            $= 4.5556 *10^ {7} m/s$

For the electron to move in a circular path the magnetic force[$F = B q v$] must be equal to the centripetal force[$\frac{mv^2}{r}$] and this is mathematically represented as

                  $Bqv = \frac{mv^2}{r}$

making B the subject

                $B = \frac{mv}{rq}$

r is the radius with a value = 5.4cm = $= \frac{5.4}{100} = 5.4*10^{-2} m$

Substituting values

                $B = \frac{9.1039 *10^{-31} * 4.556 *10^7}{5.4*10^-2 * 1.60218*10^{-19}}$

                     $= 0.0048 T$