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muminat
3 years ago
12

A proton moves with a speed of 1.17 105 m/s through Earth's magnetic field, which has a value of 50.0 µT at a particular locatio

n. When the proton moves eastward, the magnetic force acting on it is directed straight upward, and when it moves northward, no magnetic force acts on it.
(a) What is the direction of the magnetic field?

(b) What is the strength of the magnetic force when the proton moves eastward? ?N

(c) Calculate the gravitational force on the proton and compare it with the magnetic force. Compare it also with the electric force if there were an electric field with magnitude E = 1.50 102 N/C at that location, a common value at Earth's surface. Note that the mass of the proton is 1.67 10-27 kg. Fgrav = N Felec = N

Suppose an electron is moving due west in the same magnetic field as in the example at a speed of 2.95 105 m/s. Find the magnitude and direction of the magnetic force on the electron. F = ?N
Physics
1 answer:
vlabodo [156]3 years ago
7 0

a) Southward you need to apply right hand rule. If you close your hand to the east, your thumb will indicate south.

b) Given the equation for Magnetic Force

F= qVB

Replacing

F= (1.16*10^{-19})(1.17*10^5)(50*10^{-6})

F=9.36*10^{-19}

c) Given the second Newton's Law by

F_g = 1.67*10^{-27}*9.81

F_g = 1.64*10^{-26}

Given the electric force by,

F_e = 1.6*10^{-19}*1.5*10^2

F_e = 2.4*10^{-17}N

F=9.36*10^{-19}N

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Answer:

Explanation:

radius of the solenoid, r = 0.05 m

length of the solenoid, l = 0.39 m

Magnetic field of the solenoid, B = 2 x 10^-5 T

Number of turns, N = 200

The magnetic field of the solenoid is given by

B=\mu _{0}ni

where, i be the current and n be the number of turns per unit length

n = N / l = 200 / 0.39 = 512.8

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Bill leaves his 60 W desk lamp on every day, including weekends, for eight hours. After one month (30 days), how much total ener
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' W ' is the symbol for 'Watt' ... the unit of power equal to 1 joule/second.

That's all the physics we need to know to answer this question.
The rest is just arithmetic.

(60 joules/sec) · (30 days) · (8 hours/day) · (3600 sec/hour)

= (60 · 30 · 8 · 3600) (joule · day · hour · sec) / (sec · day · hour)

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__________________________________

Wait a minute !  Hold up !  Hee haw !  Whoa ! 
Excuse me.  That will never do.
I see they want the answer in units of kilowatt-hours (kWh).
In that case, it's

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14 kWh

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