Question

An electron passes through a point 2.83 cm 2.83 cm from a long straight wire as it moves at 35.5 % 35.5% of the speed of light perpendicularly toward the wire. At that moment a switch is flipped, causing a current of 17.7 A 17.7 A to flow in the wire. Find the magnitude of the electron's acceleration a a at that moment.

Answer 1

Answer:

The magnitude of electron acceleration is $2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

Explanation:

Given:

Distance from the wire to the field point $r = 2.83 \times 10^{-2}$ m

Speed of electron $v = 35.5 \%c$

Current $I = 17.7$ A

For finding the acceleration,

First find the magnetic field due to wire,

  $B = \frac{\mu _{o}I }{2\pi r }$

Where $\mu_{o} = 4\pi \times 10^{-7}$

  $B = \frac{4\pi \times 10^{-7} \times 17.7 }{2\pi (2.83 \times 10^{-2} ) }$

  $B = 12.50 \times 10^{-5}$ T

The magnetic force exerted on the electron passing through straight wire,

  $F = qvB$  

  $F = 1.6 \times 10^{-19} \times 0.355 \times 3 \times 10^{8} \times 12.50 \times 10^{-5}$

  $F = 21.3 \times 10^{-16}$ N

From the newton's second law

  $F = ma$

Where $m =$ mass of electron $= 9.1 \times 10^{-31}$ kg

So acceleration is given by,

   $a = \frac{F}{m}$

   $a = \frac{21.3 \times 10^{-16} }{9.1 \times 10^{-31} }$

   $a = 2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

Therefore, the magnitude of electron acceleration is $2.34 \times 10^{15}$ $\frac{m}{s^{2} }$