- #1

- 452

- 0

## Homework Statement

A bar magnet experiences a torque of magnitude 0.075 Nm when it is perpendicular to a 0.50 T external magnetic field. What is the strength of the bar magnet's on-axis magnetic field at a point 20 cm from the center of the magnet?

## Homework Equations

[tex]\tau = \vec{\mu} B sin\vartheta[/tex]

[tex]B = \frac{\mu_0}{4 \pi} \frac{2 \vec{\mu}}{z^3}[/tex]

## The Attempt at a Solution

since its perpendicular, sin90 = 1 so:

[tex]\tau = \vec{\mu} B \Rightarrow \vec{\mu} = \frac{\tau}{B} = \frac{0.075}{0.5} = 0.15[/tex]

[tex]B = \frac{\mu_0}{4 \pi} \frac{2 \vec{\mu}}{z^3} = 10^-7 \frac{2(0.15)}{0.2^3} = 37.5 T[/tex]

Any help would be appreciated. Thank You.