An insulated wire with mass 5.80×10 5 kg is bent into the shape of an inverted U

Question

An insulated wire with mass 5.80×105 kg is bent into the shape of an inverted U such that the horizontal part has a length l = 15.7 cm . The bent ends of the wire are partially immersed in two pools of mercury, with 3.0cm of each end below the mercury's surface. The entire structure is in a region containing a uniform 7.00×103-T magnetic field directed into the page (see the figure (Figure 1) ). 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 36.0 cm into the air, measured from its initial position.

Determine the speed v of the wire as it leaves the mercury.

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.

Ignoring the resistance of the mercury and the circuit wires, determine the resistance of the moving wire.

Explanation / Answer

After the wire leaves the mercury its acceleration is g, downward. The wire travels upward a total distance of 0.360 m

from its initial position. Its ends lose contact with the mercury after the wire has traveled 0.03 m, so the wire travels

upward 0.33 m after it leaves the mercury .

a = -9.8 m/s62

vy^2 = uy^2 + 2a*s

vy = 0 (at maximum height)

s = 0.33

uy = sqrt(2g*0.33)

uy = 2.54 m/s

part b )

Now consider the motion of the wire while it is in contact with the mercury

s = 0.03 m

initiall velocity = 0 final velocity = 2.54 m/s

vy^2 = uy^2 + 2ay*s

ay = 107.8 m/s^2

sum of all forces in y direction = may

Fb - mg = may

Fb = ILB

ILB = m(g+ay)

I = m(g+ay)/LB

L = 15.7 cm = 0.157 m

m = 5.8 x 10^-5 kg

B = 7 x 10^-3 T

I = 6.206 A

part c )

R = V/I = 15/6.206

R = 0.24 ohm

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