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:$I=112.094\ A$

Explanation:

Given

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

diameter $d=0.3\ mm$

Magnetic field $B=5.5\times 10^{-5}\ T$

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

$\rho ALg=BIL\times \sin 90$

$8900\times \frac{\pi}{4}(0.3\times 10^{-3})^2L\times 9.8=5.5\times 10^{-5}\times I\times L$

$I=\frac{8900\times \pi\times 9\times 10^{-8}\times 9.8}{4\times 5.5\times 10^{-5}}$

$I=112.094\ A$