Question

A copper wire (density equal to 8900 kg / m3) 0.3 mm in diameter "floats" due to the force of the earth's magnetic field, which is horizontal, perpendicular to the wire, and 5.5x10-5 T in magnitude. What current should the wire carry?

Answer 1

Answer:
`I=112.094\ A`

Step-by-step explanation:

Given

density
`\rho =8900\ kg/m^3`

diameter
`d=0.3\ mm`

Magnetic field
`B=5.5* 10^(-5)\ T`

Force on the current carrying conductor placed in a magnetic field

`F=BIL\sin \theta`

where L=length of conductor

`\theta`=angle between magnetic field and current

If the wire is floating then weight must be balanced by weight of wire

`Weight=mg=\rho ALg`

Therefore

`\rho ALg=BIL* \sin 90`

`8900* (\pi)/(4)(0.3* 10^(-3))^2L* 9.8=5.5* 10^(-5)* I* L`

`I=(8900* \pi* 9* 10^(-8)* 9.8)/(4* 5.5* 10^(-5))`

`I=112.094\ A`