Levitating wire!

Electricity and Magnetism Level 3

When a current carrying wire is placed in uniform magnetic field, it experiences a force. The magnitude and direction of the force $F$ depends on the magnitude and direction of current $i$ and magnetic field $B$ . $\vec{F} = i \vec{l} \times \vec{B}$ Here, $l$ is the length of the wire.
A set up can be designed in which this magnetic force can be made to oppose the gravitational force and if enough current is flown through it then it may levitate in air.

A wire of length $1\text{ m}$ and mass $10 \text{ grams}$ is placed along the equator of the Earth. The magnetic field at the location of the wire is $\SI{50}{\micro\tesla}$ and is pointing towards the North. What should be the minimum current in the wire so that it levitates?

1000 A

towards East

2000 A

towards East

2000 A

towards West

1000 A

towards West

1 solution

Tom Engelsman
Jan 25, 2017

Let us take the unit vectors:

$\begin{aligned} \text{East/West} &= \pm \hat{i}; \\ \text{North/South} &= \pm \hat{j}; \\ \text{Into/Away from Earth} &= \mp \hat{k} \end{aligned}.$

If the wire is gravitated $-mg \hat{k}$ Newtons towards the Earth's center, then we require the magnetic force, $\vec{F} = i \vec{l} \times \vec{B},$ to equal $mg \hat{k}$ Newtons outward.

Thus, $\vec{F} = i \vec{l} \times \vec{B} = il \hat{i} \times B \hat{j} = mg \hat{k}$ , Or, $\begin{aligned} ilB &= mg \\ i &= \frac{mg}{lB} \\ &= \frac {(0.010 kg)(10 m/s^{2})}{(1 m)(50 \times 10^{-6} T)} \\ i &= \boxed{2000 A Eastward}. \end{aligned}$

Levitating wire!

1 solution

0 pending reports