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dbambery
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Homework Statement
First part of the question:
A rectangular loop is placed near a long straight conductor.
Conductor is oriented vertically to the XY plain at the origin the loop is contained within the YZ plain.
Current passing through conductor.
I = 19 Ampere
dimensions of the loop
height = 9cm
length = 14cm
distance form the conductor = 5cm
Second part of the question : (The loop is assumed to have a break in it to measure induced emf.)
The loop is moved at a velocity of 2m/s in the positive Y direction.
calculate the emf across the break in the loop.
Third part of the question:
A 100 Ohm resistor is placed in the break of the loop, after 10 millisecond of motion what will be the quantity of current through the resistor.
Homework Equations
Magnetic field:
B= [\mu_{0}*I]/[2*\Pi*R]
Magnetic flux:
\Phi= integral {B.dA} over the surface (B.dA is a dot product).
The Attempt at a Solution
Solution for the first part:
As the magnetic field decays over the area to be calculated i decided to integrate the field between the 2 radii that represented the start of the coil and the end of the coil.
getting B = \int(BdR) from 0.05 lower to 0.19 upper limit.
taking out the non affected terms i got:
B= ([(mu0)*I]/[2*\Pi])\int(dR/R) from 0.05 lower to 0.19 upper limit.
further solved into:
B= ([(mu0)*I]/[2*\Pi])ln(.19/.05) = ([(mu0)*I]/[2*pi])*(1.335)
Inserting the values i got around:
B= 5.073 * 10^(-6) Tesla.
Since the field is tangential in action it can be treated as perpendicular to the surface.
\Phi = B*A
which evaluated to around 63.919 * 10^(-9) Webber.
Is the method correct I do not think that I have violated any laws, my concern being that this may cascade into the other parts so I would like a second opinion.
Second part of the question:
Differentiating the flux \Phi with respect to time gives us the induced emf.
In the case of this question the magnetic field does not change rather the area through which the magnetic lines are passing.
\Phi = B*A
\Phi = B*L*x L here being the lateral dimension of the loop ie the height and the x being the length of the loop contained within the field.
As previously stated the rate of change of x is known to be 2m/s.
when differentiating the equation;
\Phi = B*L*x wrt time we get
\Phi = B*L*(\frac{dx}{dt})
Putting in the values provided i get:
0.913 * 10^(-6) Volt.
Third section:
In this section should i re calculate the magnetic flux for the new position and then insert that into my previous equation to get the emf in the loop at that point and then apply Ohm's law.
Sorry guys this is my first post, apologies in advance for any inconvenience caused.
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