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natulia [17]
3 years ago
5

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.
Physics
2 answers:
marusya05 [52]3 years ago
7 0

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

dimaraw [331]3 years ago
4 0

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}

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