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boyakko [2]
3 years ago
14

A long, straight wire lies in the plane of a circular coil with a radius of 0.018 m. the wire carries a current of 5.6 a and is

placed along a diameter of the coil. (a) what is the net flux through the coil? (b) if the wire passes through the center of the coil and is perpendicular to the plane of the coil, find the net flux through the coil.
Physics
1 answer:
iris [78.8K]3 years ago
5 0
(a) The net flux through the coil is zero.
In fact, the magnetic field generated by the wire forms concentric circles around the wire. The wire is placed along the diameter of the coil, so we can imagine as it divides the  coil into two emisphere. Therefore, the magnetic field of the wire is perpendicular to the plane of the coil, but the direction of the field is opposite in the two emispheres. Since the two emispheres have same area, then the magnetic fluxes in the two emispheres are equal but opposite in sign, and so they cancel out when summing them together to find the net flux.

(b) If the wire passes through the center of the coil but it is perpendicular to the plane of the wire, the net flux through the coil is still zero.
In fact, the magnetic field generated by the wire forms concentric lines around the wire, so it is parallel to the plane of the coil. But the flux is equal to
\Phi = BA \cos \theta
where \theta is the angle between the direction of the magnetic field and the perpendicular to the plane of the coil, so in this case \theta=90^{\circ} and so the cosine is zero, therefore the net flux is zero.
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Where B = magnetic field due to long wires, \mu_{0} = 4\pi \times10^{-7}, R = perpendicular distance from wire to given point

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∴ B = B_{1} + B_{2}

 B = \frac{\mu_{0} i}{2\pi R_{1} } +  \frac{\mu_{0} i}{2\pi R_{2} }

 B =\frac{ 4\pi \times10^{-7} \times5}{2\pi } [\frac{1}{0.03} + \frac{1}{0.05} ]

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