Gaussmeter wire electric field

[weijey1]

A simple gaussmeter for measuring horizontal magnetic fields consists of a stiff 50 cm wire that hangs from a conducting pivot so that its free end makes contact with a pool of mercury in a dish below. The mercury provides an electrical contact without constraining the movement of the wire. The wire has a mass of 1 g and conducts a current downward.

(a) What is the equilibrium angular displacement of the wire from vertical if the horizontal magnetic field is 0.04 T and the current is 0.20 A?


(b) If the current is 20 A and a displacement from vertical of 0.5 mm can be detected for the free end, what is the horizontal magnetic field sensitivity of this gaussmeter?

 

[member 553083]

(a) The equilibrium angular displacement of the wire from vertical due to the magnetic field is given by the equation $\theta = \frac{\mu_0 I L B}{2 mg}$, where $\mu_0$ is the permeability of free space, $I$ is the current, $L$ is the length of the wire, $B$ is the magnetic field, and $m$ is the mass of the wire. Substituting the given values into the equation gives $\theta = \frac{(4 \pi \times 10^{-7})(0.2 \text{ A})(0.5 \text{ m})(0.04 \text{ T})}{2 (1 \text{ g})} = 1.26 \times 10^{-4} \text{ radians}$. (b) The sensitivity of the gaussmeter can be determined from the equation $\frac{\Delta B}{B} = \frac{\Delta \theta}{\theta}$, where $\Delta B$ is the change in the magnetic field, $B$ is the magnetic field, $\Delta \theta$ is the change in the angular displacement of the wire, and $\theta$ is the angular displacement of the wire. Substituting the given values gives $\frac{\Delta B}{B} = \frac{0.5 \times 10^{-3} \text{ m}}{1.26 \times 10^{-4} \text{ radians}} = 3.97 \times 10^3 \text{ T}^{-1}$. Therefore, the sensitivity of the gaussmeter is 3.97 mT/A.