Question

An electron passes through a point 3.47 cm from a long straight wire as it moves at 30.5 % of the speed of light perpendicularly toward the wire. At that moment a switch is flipped, causing a current of 12.7 A to flow in the wire. Find the magnitude of the electron's acceleration a at that moment.

Answer 1

Answer:

the magnitude of the electron's acceleration a at that moment = $1.176*10^{15} \ m/s^2$

Explanation:

The expression for the magnetic field is :

$B = \frac{\mu_0 I}{2 \pi r}$

where:

$\mu_0$ = permeability of free space = $4 \pi * 10^{-7} \ T . m/A$

I = current in the wire = 12.7 A

r = distance from the wire to the field point = 3.47  cm = $3.47 *10^{-2} m$

$B = \frac{(4 \pi *10^{-7}(12.7)}{2 \pi (3.47*10^{-2})}$

$B = 7.3199*10^{-5} \ T$

The magnetic force exerted in the electrond passing through the straight conducting wire is:

F = Bvgsinθ

where :

B = magnetic field = $7.3199*10^{-5} \ T$

v = speed of light = $(3*10^8 \ m/s)$(30.5 % )

q = charge of electrons = $1.6*10^{-19} \ C$

θ = angle between speed of electron & field = 90°

F = ($7.3199*10^{-5} \ T$) × $(3*10^8 \ m/s)$(30.5 % ) × ($1.6*10^{-19} \ C$) × ( sin 90°)

F = $1.0716 *10^{-15} \ N$

From Newtons second Law ;

F = ma

$a = \frac{F}{m}$

where m = mass = $9.11*10^{-31} \ kg$

$a = \frac{1.0716*10^{-15 }N}{9.11*10^{-31} \ kg}$

$a = 1.176*10^{15} \ m/s^2$

Thus, the magnitude of the electron's acceleration a at that moment = $1.176*10^{15} \ m/s^2$

Answer 2

Answer:

The acceleration of the particle is 1.17x10¹⁵m/s²

Explanation:

The equation of the magnetic field is equal to:

$B=\frac{u_{o}I }{2\pi r}$

Where

uo = permeability of free space = 4πx10⁻⁷Tm/A

I = current = 12.7 A

r = distance from the wire = 3.47 cm = 0.0347 m

$B=\frac{4\pi x10^{-7}*12.7 }{2\pi *0.0347} =7.32x10^{-5} T$

The magnetic force is:

$F=Bvqsin\theta$

Where

v = speed of electron = 0.305 * 3x10⁸ = 9.15x10⁷m/s

q = charge of electron = 1.6x10⁻¹⁹C

θ = 90°

$F=7.32x10^{-5} *9.15x10^{7} *1.6x10^{-19} *sin90=1.07x10^{-15} N$

The acceleration of the particle is

$a=\frac{F}{m} =\frac{1.07x10^{-15} }{9.11x10^{-31} } =1.17x10^{15} m/s^{2}$