A simple gaussmeter for measuring horizontal magnetic fields consists of a stiff 52 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 3 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?

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opudodennis opudodennis · Answer 1 · 2020-04-01T15:03:48+02:00

Answer:

8.1345°

Explanation:

We apply [tex]\sum F_x=0[/tex] to the wire to obtain:

[tex]\sum F_x=-F_m+mg\ sin \theta=0[/tex]

#The magnitude of the magnetic force acting on the wire is given by:

[tex]F_m=IlB\ sin \phi, \phi=90\textdegree\\\\F_m=IlB[/tex]

#Substitute for [tex]F_m[/tex] to obtain:

[tex]-IlB+mg\ sin \ \theta=0\ \ \ \ \ \ \ \ ...i[/tex]

Solve for [tex]\theta[/tex]:

[tex]\theta=sin^{-1}[\frac{IlB}{mg}][/tex]

We the substitute the numerical values to calculate the equilibrium angular displacement:

[tex]\theta= sin^{-1}[\frac{0.2A\times0.52m\times 0.040T}{0.003\ kg\times9.8m/s^2}]\\\\=8.1345\textdegree[/tex]

Hence, the equilibrium angular displacement of the wire from vertical if the horizontal magnetic field is 8.1345°