Tangent galvanometer reversal and residual field
In an experiment to measure the horizontal component of the Earth's magnetic field, a tangent galvanometer has 300 turns and a mean radius of 10.0 cm. The coil plane is set in the magnetic meridian, and the needle is checked to swing freely. A current of 20.0 mA is passed east-to-west and the needle settles at 45.0° to the meridian. On reversing the current, the needle settles at 43.0° on the other side. The student computes B = μ0nI/(2r), obtains 37.7 μT, and reports that as the horizontal component of the Earth's field. The student says the 2° difference is pivot friction and ignores it. The apparatus can be rotated about a vertical axis, and the current can be varied with a rheostat.
1. Calculate the coil field and use both deflections to find the horizontal component and the residual field along the coil axis.
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Bc = μ0nI/(2r) = (4π×10^-7)(300)(0.0200)/(2×0.100) = 3.77×10^-5 T = 37.7 μT. Let Br be the residual field along the coil axis (east-west). For the two current directions, tan θ = (Bc ± Br)/Bh. Thus Bc + Br = Bh tan45° = Bh, and Bc - Br = Bh tan43° = 0.9325 Bh. Adding: 2Bc = Bh(1 + 0.9325), so Bh = 75.4/1.9325 = 39.0 μT. Then Br = Bh - Bc = 39.0 - 37.7 = 1.3 μT. The sign of Br depends on which current direction is called positive; its magnitude is about 1.3 μT.
2. Critique the friction explanation and describe one test that distinguishes friction from a residual field.
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Friction at the pivot would not produce a consistent 2° shift that reverses with the current. A residual field adds vectorially to Bc and therefore changes tanθ differently for the two current directions. To test, pass no current and see whether the needle points exactly along the magnetic meridian; a residual field will deflect it. Alternatively, rotate the coil through 180° about a vertical axis: the residual-field effect will change sign relative to the coil, while a pivot bias will not behave in the same way. Repeating at several currents and plotting tanθ against I will give a line whose intercept is Br/Bh.
3. The student proposes to increase the current to 100 mA to reduce the residual-field error. Will this help, and what practical limits appear?
Read the worked answer
Increasing the current increases Bc and drives the deflection toward 90°, where a small field difference produces a very small angle change. At 100 mA, Bc = 5 × 37.7 = 188.5 μT. With Bh = 39.0 μT and Br = 1.3 μT, tanθ1 = (188.5 + 1.3)/39.0 = 4.866, θ1 = 78.36°, and tanθ2 = (188.5 - 1.3)/39.0 = 4.800, θ2 = 78.24°. The difference is only 0.12°, so the residual-field error is much reduced. However, the needle is then close to 90°, where reading precision is poor, and the coil may heat; a moderate current with reversal and averaging is better.