Notes About Electricity and Magnetism

7/29/2019 Notes About Electricity and Magnetism

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Electric field due to a line of charge

Find the electric field at a point due to a line of charge. See figure 1.

We first position the point where the electric field is intended to be measured at the origin, see

figure 2.

Each segment of the line contributes to the total field by:

(Note that (distance) is different for each segment). Thus, the total electric field will be:

In order to integrate, we must have in terms ofR, therefore:

[ ] []

[ ] [ ]


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If the point of interest is not located on the x-axis (at the same level as the line of charge) as in the

figure 3 below

Then:

Where:

Then:


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* + * + * +* +

Electric field due to a semi-circle ring of charge (figure 4)


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Electric field due to a ring of charge (figure 5)

( )


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Note that:

Which is the same as the electric field due to a point of charge.

If instead of a ring of charge we had a uniformly charged disk (figure 6) so that:

Then, we can consider the disk as an infinitesimal number of rings with radii extending from 0 to R,

with each ring carrying a charge of. Thus:


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[ ]

[ ]Using our knowledge of

we can write the previous equation as:

[ ] [ ]

Note that:

[

]

[ ]

Thus if the distance from the centre of the disk is very small in comparison to the radius or, the disk

is very large in comparison to the distance from the disk then:

The last equation is valid not only for a disk but, for any shape flat plate as long as the distance to

the point of interest is very small in comparison to the distance to the edge of the plate (

).


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Torque due to a circular ring of current (figure 7)

If we assumed that the same time is required for the current to travel from point 1 to point 2, then:

However, normally only the current () is known and not the vertical component , thus we shouldassume that the velocities and are equal (however travelling different distances) to obtain thetime () required for the current to travel the distance of, hence:


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* + * +


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Magnetic Field due to a wire of current (figure 8)

[ ]


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Note that when:

This means when the wire length is significantly longer that the radius (which is the case with most ifnot all wires), then the magnetic field is:

Deriving RMS value of a sinusoidal current

The graph of AC current and voltage is a sine graph:

Where:

: Peak voltage: is Angular velocity in radian/seconds .: is time (period) in seconds.Note that

is the angle in radians that the coil is making as it rotates over time.

For a constant non-reactive resistance the power against time (period) graph however, will be the

square of a sine graph (see figure 9):


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Note that because is assumed to be constant (rpm of generator is constant, this also means thatthe frequency of the current is constant), therefore in addition to power the only other variable in

this function is time (period), hence the graph of the power equation was taken against time only,

where:

To obtain the average power the sum of the tangent of the graph (area under the graph) is

calculated and divided over the period (area under the curve = (W.s = J), dividing by (s) givesP (W)), thus:

Using a trigonometric identity to eliminate squaring of trigonometry function:

* +

* +Note that thus:

*

+

[ ] [ ] [ ]
http://en.wikipedia.org/wiki/List_of_trigonometric_identitieshttp://en.wikipedia.org/wiki/List_of_trigonometric_identities


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Note that we could have used the instead, to derive the power in terms of voltage.Moreover, using the RMS power equation we can easily calculate and in terms of and:

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Notes About Electricity and Magnetism