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pc2-brazil
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Homework Statement
At a certain place, the Earth's magnetic field has a magnitude B = 59 μT and is inclined downward at an angle of 70º to the horizontal. A flat horizontal circular coil of wire with a radius of 13 cm has 950 turns and a total resistance of 85 Ω. It is connected to a galvanometer with 140 Ω resistance. The coil is flipped through a half revolution about a diameter, so it is again horizontal. How much charge flows through the galvanometer during the flip? (Hint: See Problem 11.)
Homework Equations
The "Problem 11" referred to in the question concludes that, if a circular loop is connected to a resistance R, this configuration is perpendicular to a magnetic field B, Φ(0) is the flux in the loop at time t = 0, and the magnetic field B varies in a continuous but not specified way, so that at time t the flux is represented by ΦB(t), then the net charge q(t) that has passed through resistor R in time t is:
[tex]q(t)=\frac{1}{R}[\Phi_B(0)-\Phi_B(t)][/tex]
The Attempt at a Solution
If the circular coil is horizontal, and Earth's magnetic field at that point is inclined downward at an angle of 70º to the horizontal, then angle between the plane of the coil and Earth's magnetic field is 70º. So, if we adopt the normal to the plane of the coil to be downward, then the angle θ between the normal to the plane and the magnetic field is 160º. So, in its initial position, the magnetic flux through the surface of the N loops should be:
[tex]\Phi(0)=NAB\cos{160º}[/tex],
where [tex]A=\pi r^2[/tex] is the area of a loop.
Thus, [itex]\Phi(t)[/itex] would be the negative of the above, since the loop is 180º from its original inclination.
Also, I think that the resistance R is 140 Ω + 85 Ω = 225 Ω.
So, the result would be:
[tex]q(t)=\frac{1}{R}[NAB\cos{160º}-(-NAB\cos{160º})] = \frac{1}{R}[2NAB\cos{160º})][/tex]
where R = 85+140 = 225 Ω, A = 13 cm, N = 950, A = π*13² cm², B = 0.000059 T.
This gives approximately 24.86 μC.
Does this reasoning seem correct?
Thank you in advance.