(c19p66) A straight wire of mass 11.00 g and length 5.00 cm is suspended from tw
ID: 1428824 • Letter: #
Question
(c19p66) A straight wire of mass 11.00 g and length 5.00 cm is suspended from two identical springs that, in turn, form a closed circuit (Fig. P19.66). The springs stretch a distance of 0.90 cm under the weight of the wire. The circuit has a total resistance of 11.00 ?. When a magnetic field is turned on, directed out of the page (indicated by the dots in Fig. P19.66), 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.)
24 V -5.0 cm- © 2003 Thomson-Brooks Cole Figure P19.66Explanation / Answer
F = B*I*L*sin A
Sin A = 1 as A = 90 deg
I = V/R = 24/11 = 2.181 amp
L = 5 cm = 0.05 m
Fi = mg = kx
k = mg/x = 11*10^-3*9.81/0.009 = 11.99
Ft = k*dx = 11.99*0.3*10^-3 = 0.003597 N
Now
B = F/(I*L*sin A) = 0.003597/(2.181*0.05*1) = 0.0329 = 0.033 T
Comment below if you have any doubt.
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