Example: Magnetic Field

Problem

A Metal Rod with length 1.2 m and mass 500 g is suspended in a magnetic field of 0.9 T. Determine the current needed suspend the rod without supports.

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Solution

Ampere's Force Law

$\overrightarrow{F}=\underset{c}{\int }I\overrightarrow{d}l\times \overrightarrow{B}$

For our problem we have $\overrightarrow{B}=0.9T$

And we know the necessary Force to hold up the bar without the supports would be equal to the weight of the bar multiplied by the gravitational force.

We also know that the force vector will be in the $\hat{k}$ direction, and B is in the $\hat{i}$ direction.

So using the fact that $\hat{k}=-\hat{j}\times \hat{i}$ We can conclude that the current is in the $-\hat{j}$ direction, and therefore flows to the left.

${\overrightarrow{F}}_{\text{needed}}=\text{mass}\times \text{gravity}$

$\overrightarrow{F}=500*9.81(g*m/s^2)$

$\overrightarrow{F}=4.905(N)$

Substituting into Ampere's Law we are left with

$4.905=\underset{c}{\int }I\overrightarrow{d}l\times 0.9(T/N)$

Integrating from $l=0tol=1.2m$ gives us

$4.905=I*1.2*0.9(T*m/N)⟹I=4.905/(1.2*0.9)=4.5417A$

In conclusion we would need 4.54 Amps of current flowing to the left to suspend the bar in the magnetic field.

$I=4.54\hat{i}A$

Review Comments and Status

Andrew Sell- emailed 1.25.10

Kyle Lafferty- emailed 1.25.10