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course Phy 202
4/18 11 amProblem Number 1
A charge of 78 `microC is surrounded by an electric field of 36 volts / meter. Determine the potential gradient dV/dx of the field, the force exerted on the charge, the work done by the field in moving this charge .16 meters in the direction of the field, and the force per unit charge exerted by the field.
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Electric field=36 V/m q=78 microC Determine potential gradient:
The potential gradient s 36 volts/meter.
F= 78 microC * 36 volts/meter=.00078 C*36 volts/meter=0.02808 N
Work done=0.02808 N*.16 meters=.0045 J
The force per unit of Charge is 36 volts/C
@& To get the force per unit of charge you would divide .02808 N by 78 microCoulombs. You would get 36 N / C.
That's the same as 36 V / m, so the result should be no surprise.*@
Problem set electricity 1/17
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Problem Number 2
Under certain conditions of temperature and potential difference, a uniform wire with cross-sectional radius 3.6 cm carries a current of 1.7 amps. Under the same conditions, what will be the current of a second wire whose length is identical to that of the first, but which has diameter 6.3 cm?
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.V/.0036 m and V/.0063 m so the potential gradient is V/.0063 m /V/.0036 m= .0036m/.0063 m=.57 times greater than the first wire
Therefore the current is .57 *1.7 amps=.969 amps
@& Potential gradient is change in potential / distance. In this case the potential gradient would be the voltage divided by 3.6 cm. Both wires have the same length and therefore the same potential gradient.
The potential gradient is the force per unit of charge. It determines how much force is accelerating the electrons. This acceleration is the same in both wires.
The difference is in the cross-sectional areas. The more cross-sectional area, the more electrons are available to be accelerated, and the greater the current.
The question is therefore how the cross-sectional areas of the two wires compare. *@
Problem set 2/2
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• . ratio of potential gradients: (V / L2) / (V / L1) = L1 / L2.
Problem Number 3
A coil consists of 36 circular loops of wire each of radius .19 meters. The coil is in a uniform magnetic field with strength .0077 Tesla. The coil is rotated in such a way that at t = 0 the field makes an angle of 53 degrees with a perpendicular to the plane of the loop, while at t = .01 sec the field makes an angle of -9.001 degrees with the same perpendicular. What is the average voltage induced around the coil?
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radius=.19 m
36 loops
B=.0077 Tesla Angle 53 degrees with perpendicular, at t=0
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@& Flux = field * area when the field is perpendicular to the region.
If the field isn't perpendicular to the region, the component of the field perpendicular to the region is used to find the flux.
In this case the component perpendicular to the region is .0077 Tesla * cos(53 deg).
This would be multiplied by the total area of the coil, which is 19 times the area of a single circular loop.*@
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Problem Number 4
A uniform electric field of strength 50000 N/C exists between the plates of a capacitor. The plates are 1.5 cm apart. If 1.12 Joules of work are required to move a charge from one plate to the other, what is the charge?
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W=F*`dx
F=W/`dx=1.12 J/.015 m=74.67 N
so q=74.67 N/50000N/C=.00149 C
@& Good.*@
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Problem Number 5
Find the strength of the magnetic field due to a straight current segment of length .043 m, at a distance of 5.9 meters from the segment, provided that the vector from the segment to the point is perpendicular to the segment, and that a current of 3.5 Amps flows in the segment.
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L=.043 m r= 5.9 m I= 3.5 Amps.
B= .0000001 Tesla / Amp meter (.043 m*3.5 Amp)/(5.9 m)^2
B== .0000001 Tesla / Amp (.1505 Amp m) /32.45 m^2= 4.64 x 10^-10 Tesla
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@& Good.*@
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Problem Number 6
A sensitive balance shows that a force of .0087 Newtons is exerted on a straight wire 3 cm long carrying a current of 5.7 amps by a uniform magnetic field directed perpendicular to the wire. What is the strength of the field? If the force is to the East and the current runs to the South, what is the direction of the field?
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F=.0087 N x=.03 m current 5.7 amps
B=9 * 10^-7 T m / amp(5.7 amps*.03m/.015 m^2)
I know this is not completely right. How do I find the field strength without an angle, or is the angle 90 degrees?
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@& The field is perpendicular to the wire, so the angle is 90 degrees.
The right-hand rule determines the direction of the field. The field and the current are perpendicular, and the force is perpendicular to both. So the field is perpendicular to the force (and, as is given, to the current).
What directions are therefore candidates for the field?
Which of these directions is confirmed by the right-hand rule?*@
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Problem Number 7
A certain hypothetical atom contains 67 protons and 86 neutrons in its nucleus and has an atomic mass of 151.9 atomic mass units, or amu (an amu is approximately 1.66 * 10^-27 kg). How many protons and how many neutrons will it end up with if it undergoes an alpha decay? How many if it undergoes a beta decay? How many if it undergoes a gamma decay?
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This is supposed to be part of this test?
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@& No. There are some renegade problems on the test.*@
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Problem Number 8
A generator has a negligible internal resistance. It creates a potential difference of 5.5 volts when cranked at 1 revolution/sec, and the voltage is proportional to the cranking rate. The generator cranked at 1.8 revolutions / sec, and is connected to a series combination of two bulbs, one with a resistance of 5.5 ohms and the other with resistance 51 ohms. How much current flows in each bulb? How much power is dissipated in each?
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I=V/R
I1=5.5 V/5.5 ohms=1 amp
12=5.5 V/51 ohms=.108 amp
Power1= 1C.s*5.5 J/C=5.5 watts
Power2=.108 C/s*5.5J/C=.594 watts
@& Two resistances in series will not both experience the full voltage of the source. The sum of the voltage drops will equal the voltage of the source.
If the bulbs were in parallel your solution would be correct for a voltage of 5.5 volts.
However the cranking rate implies a voltage of 1.8 * 5.5 volts.*@
@& Check my notes. I'll be glad to answer more questions.*@