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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.

Respuesta :

ajeigbeibraheem

Answer:

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

Explanation:

The expression for the magnetic field is :

[tex]B = \frac{\mu_0 I}{2 \pi r}[/tex]

where:

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

I = current in the wire = 12.7 A

r = distance from the wire to the field point = 3.47  cm = [tex]3.47 *10^{-2} m[/tex]

[tex]B = \frac{(4 \pi *10^{-7}(12.7)}{2 \pi (3.47*10^{-2})}[/tex]

[tex]B = 7.3199*10^{-5} \ T[/tex]

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

F = Bvgsinθ

where :

B = magnetic field = [tex]7.3199*10^{-5} \ T[/tex]

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

q = charge of electrons = [tex]1.6*10^{-19} \ C[/tex]

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

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

F = [tex]1.0716 *10^{-15} \ N[/tex]

From Newtons second Law ;

F = ma

[tex]a = \frac{F}{m}[/tex]

where m = mass = [tex]9.11*10^{-31} \ kg[/tex]

[tex]a = \frac{1.0716*10^{-15 }N}{9.11*10^{-31} \ kg}[/tex]

[tex]a = 1.176*10^{15} \ m/s^2[/tex]

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

onyebuchinnaji

The acceleration of the electron at that moment is 1.178 x 10¹⁵ m/s².

The given parameters;

  • position of the electron, r = 3.47 cm
  • speed of light, c = 3 x 10⁸ m/s
  • current flowing in the field, I = 12.7 A

The magnitude of the magnetic field is calculated as follows;

[tex]B = \frac{\mu_0 I}{2\pi r} \\\\B = \frac{(4\pi \times 10^{-7}) \times (12.7)}{2\pi \times (3.47\times 10^{-2})} \\\\B = 7.32 \times 10^{-5} \ T[/tex]

The magnetic force on the electron is calculated as follows;

F = qvB

[tex]F = (1.602 \times 10^{-19}) \times (0.305 \times 3\times 10^8) \times (7.32 \times 10^{-5})\\\\F = 1.073 \times 10^{-15}\ N[/tex]

The acceleration of the electron is calculated as follows;

[tex]a = \frac{F}{m} \\\\a= \frac{(1.073\times 10^{-15})}{9.11 \times 10^{-31} } \\\\a = 1.178 \times 10^{15} \ m/s^2[/tex]

Thus, the acceleration of the electron at that moment is 1.178 x 10¹⁵ m/s².

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