A straight wire of mass 10.2 g and length 5.0 cm is suspended from two identical

Question

A straight wire of mass 10.2 g and length 5.0 cm is suspended from two identical springs that, in turn, form a closed circuit. The springs stretch a distance of 0.45 cm under the weight of the wire. The circuit has a total resistance of 13 . When a magnetic field directed out of the page (indicated by the dots in the figure) is turned on, the springs are observed to stretch an additional 0.30 cm. What is the strength of the magnetic field? (The upper portion of the circuit is fixed.) ?=_______ T

Explanation / Answer

1) Find spring constant
F= -kx
k= - 0.0102kg * 9.8 m/s^2 /0.0045 m
k = 22.21 N/m

(2) Find force on wire from magnetic field
F =-k x
F = 22.21 N/m * -.003 m
F = .066 N

(3) Find the current in the loop
V=IR
R = 13 ?
V = 24 V
I = 1.85 amps

(4) Apply Lorentz Force Law

F = I L x B
L = 0.05m
I = 1.85 amp ... (3)
F = .066N ... (2)
B = .066N/(1.85A x 0.05m)

B = 0.713 Tesla

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