Question

Let's say you want to "float" a section of copper wire, which is 7.65 cm long and 2.34 mm in diameter within the magnetic field of Earth near the Earth's surface. Assume the magnetic field of Earth makes a 31° with respect to the horizontal and has a magnitude of 3.5.x10-5 T. The current is flowing from west to east. See the diagram given below. 5.Draw a Free Body Diagram of the wire. 22-degrees B 6. What is the value of current needed to suspend this section of copper wire? 7. What is the emf as a function of time produced in the wire due to its unbalanced force? How would this effect the current in the wire? Would its vertical position increase, decrease, or stay the same?

Answer 1

Length of the copper wire L=7.65 cm= 7.65*10^-2 m

Diameter of the wire=2.34 mm

Radius r=2.34/2=1.17*10^-3 m

Angle made by the magnetic field with horizontal $\theta$ =31^o

Magnetic field of earth=3.5*10^-5T

Thus, horizontal component of the magnetic field, B_V=B sin$\theta$=3.5*10^-5*sin31=1.8*10^-5T

5) The free body diagram of the current carrying wire is shown below.

6) To suspend or float a section of copper wire, the magnetic field due to the current carrying wire shorld be equal to the vertical component of the earth's magnetic field as shown above.

Bc = po I 2r I BH N. 31° Bu To

Thus, $\mu$ oI/2r=1.8*10^-5

where $\mu$ o= permeability of free space=4$\pi$*10^-7N/A²

Using r=1.17*10^-3 m(given),

Current through the wire I=1.8*10^-5*3.34*10^-3/(4$\pi$*10^-7)=0.048A

7) According to Faraday's law emf e=-d$\phi$/dt, where $\phi$ =magnetic flux=B.dA

For the wire, dA=2$\pi$rL

Thus, e=-d/dt( $\mu$ oI/2r*2$\pi$rL)=-d/dt( $\pi$ $\mu$oI/L)

This will change the current in the loop, in which emf changes according to the Lenz law in a direction opposite to the source producing it, thus increasing or decreasing the current.

The vertical position remains unchanged since the magnetic field and hence the current is independent of the vertical length L.