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3 years ago

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At the equator, near the surface of the Earth, the magnetic field is approximately 50.0 μT northward, and the electric field is

about 100 N/C downward in fair weather. Find the gravitational, electric, and magnetic forces on an electron in this environment, assuming that the electron has an instantaneous velocity of 8.50 ✕ 106 m/s directed to the east.

1 answer:

MArishka [77]3 years ago

6 0

Answer:

A) F_g = 8.9278 × 10^(-30)

B) F_e = 1.6 × 10^(-17) N upwards

C) F_m = 6.8 × 10^(-17) N downwards

Explanation:

A) Formula for gravitational force is;

F_g = m_e × g

m_e is mass of electron = 9.11 x 10^(-31) kg

F_g = 9.11 x 10^(-31) × 9.8

F_g = 8.9278 × 10^(-30) N downwards

B) Formula for Electric force is;

F_e = qe

q is charge on electron = 1.6 × 10^(-19) C

E is electric field = 100 N/C

F_e = 1.6 × 10^(-19) × 100

F_e = 1.6 × 10^(-17) N upwards

C) Magnetic force is given by the formula;

F_m = qVB

q is charge on electron = 1.6 × 10^(-19) C

V is velocity given as 8.50 × 10^(6) m/s

B is magnetic field = 50.0 μT = 50 × 10^(-6) T

F_m = 1.6 × 10^(-19) × 8.50 × 10^(6) × 50 × 10^(-6)

F_m = 6.8 × 10^(-17) N downwards

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b) $10.526 s$

c) $61.82 N$

Explanation:

<h3>a) Impulse delivered to the ball</h3>

According to the Impulse-Momentum theorem we have the following:

$I=\Delta p=p_{2}-p_{1}$ (1)

Where:

$I$ is the impulse

$\Delta p$ is the change in momentum

$p_{2}=mV_{2}$ is the final momentum of the ball with mass $m=0.685 kg$ and final velocity (to the right) $V_{2}=57 m/s$

$p_{1}=mV_{1}$ is the initial momentum of the ball with initial velocity (to the left) $V_{1}=-38 m/s$

So:

$I=\Delta p=mV_{2}-mV_{1}$ (2)

$I=\Delta p=m(V_{2}-V_{1})$ (3)

$I=\Delta p=0.685 kg (57 m/s-(-38 m/s))$ (4)

$I=\Delta p=65.075 kg m/s$ (5)

<h3>b) Time </h3>

This time can be calculated by the following equations, taking into account the ball undergoes a maximum compression of approximately $1.0 cm=0.01 m$ :

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Where:

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$a=\frac{V_{2}^{2}-V_{1}^{2}}{2d}$ (8)

$a=\frac{(57 m/s)^{2}-(-38 m/s)^{2}}{2(0.01 m)}$ (9)

$a=90.25 m/s^{2}$ (10)

Substituting (10) in (6):

$57 m/s=-38 m/s+(90.25 m/s^{2})t$ (11)

Finding $t$ :

$t=1.052 s$ (12)

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According to Newton's second law of motion, the force $F$ is proportional to the variation of momentum $\Delta p$ in time $\Delta t$ :

$F=\frac{\Delta p}{\Delta t}$ (13)

$F=\frac{65.075 kgm/s}{1.052 s}$ (14)

Finally:

$F=61.82 N$

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