The thin circular coil in the figure has radius r = 14 cm and contains N = 3 tur

Question

The thin circular coil in the figure has radius r = 14 cm and contains N = 3 turns of resistive wire. The coil has a resistance of 0.4 ohms. A small compass is placed at the center of the coil. With the battery disconnected, the compass needle points to the right, in the plane of the coil. The apparatus is located near the equator, where the Earth's magnetic field BEarth is horizontal, with a magnitude of 4×10-5 tesla. (Assume that BEarth is pointing to the right.)

(a) When the 1.4-volt battery is connected, predict the deflection of the compass needle. (If you have to make any approximations, think carefully about what they are.)
What is the magnitude of the deflection?

(b) The compass is removed. The current in the coil continues to run. An electron is at the center of the coil and is moving with speed v = 4 × 106 m/s into the page (perpendicular to the plane of the coil). In addition to the magnetic fields in the region due to the Earth and the coil, there is an electric field at the center of the coil (due to charges not shown in the figure), and this electric field points upward and has a magnitude E = 730 volts/meter. Predict the net force on the electron at this moment. (You can neglect the gravitational force, which easily can be shown to be negligible. Assume the +x axis is to the right, the +y axis is up, and the +z axis is out. Pay close attention to the orientation of the coil with respect to the Earth.)

what is the net magnetic force?

Explanation / Answer

>Whitney,

Magnetic field at the center of a coil of N turns, carrying current I=V/R, is N(μ0/4π)I.2πr/r2 = 3X(1e-7)X4X2π/0.14 = 5.386X10-5Tesla ~ 53.86 microtesla. Since current flows clock-wise, this field points in the direction perpendicular to the earth's field (into the page), and the compass aligns itself with the nett field, which is the vector sum of these two fields at right angles to each other. Hence the deflection is arctan(53.86/40) =  53.4 degrees.

Choose a right-handed coord system, x is along the earth's field, y is along the coil's B field (i.e. into the page), and z is upward. The velocity is V = 9e6 j, the electric field is E = 790 k and the magnetic field is B = 40e-6i + 53.86e-6j. Here i, j, k are unit vectors along x, y and z directions.

Lorentz equation says F=q*(E + V cross B) where q = 1.602e-19 coulomb (charge on an electron). Using j cross j = 0, and j cross i = -k  we get F = 1.6e-19*(790 - 9e6*40e-6) k=1.6e-19*(790 - 360) = 6.9X10-17 k Newtons (ans). The k means it points vertically upwards. By using full vector notation we've already "paid close attention to the coil orientation".

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