Question An insulated wire with mass m = 6.00 10-5 kg is bent into the shape of
ID: 1428251 • Letter: Q
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An insulated wire with mass m = 6.00 10-5 kg is bent into the shape of an inverted U such that the horizontal part has a length l = 10.0 cm. The bent ends of the wire are partially immersed in two pools of mercury, with 2.5 cm of each end below the mercury's surface. The entire structure is in a region containing a uniform 0.00700-T magnetic field directed into the page (see figure below). An electrical connection from the mercury pools is made through the ends of the wires. The mercury pools are connected to a 1.50-V battery and a switch S. When switch S is closed, the wire jumps 30.0 cm into the air, measured from its initial position.
(a) Determine the speed v of the wire as it leaves the mercury. answer in m/s
(b) Assuming that the current I through the wire was constant from the time the switch was closed until the wire left the mercury, determine I. answer in A
(c) Ignoring the resistance of the mercury and the circuit wires, determine the resistance of the moving wire. answer in
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1,685 answers
mass of the wire, m=6*10^-5 kg
length of the wire , l=10cm
magnetic field B=0.007 T
battery potential, V=1.5v
displacement of the wire, s=30cm
a)
use,
v^2=2*g*s
v^2=2*9.8*(30*10^-2)
====> speed v=2.42 m/sec incorrect
speed of the wire s it leaves the mercury, v=2.42 m/sec
b)
let,
current in the wire is i,
magnetic force(Fm) = gravitational forcr(Fg)
i*l*B=m*g
i*10*10^-2*0.007=6*10^-5*9.8
====> i=0.84 A
current in the wire is, i=0.84 A inccorect
c)
resistance R=V/i
R=1.5/(0.84)
R=1.78 ohms
resistance of the wire, R=1.78 ohms incorrect
Explanation / Answer
a) Here, velocity after upwards force was removed = sqrt(2 * 9.8 * 0.275)
= 2.321 m/sec
=> the speed v of the wire as it leaves the mercury = 2.321 m/sec
b) Here, net acceleration of wire in mercury = (2.321 * 2.321) / (2 * 0.025)
= 107.74 m/sec2
=> acceleration provided by magnetic force = 107.74 + 9.8
= 117.54 m/sec2
=> magnetic force = 117.54 * 6.00 *10-5
= 7.0524 * 10-3 N
=> F = I * L * B
=> 7.0524 * 10-3 = I * 0.007 * 0.1
=> I = 10.075 A ------------------> current I through the wire
c) resistance of the moving wire = 1.50 / 10.075
= 0.1488 ohm
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