# Bar Magnet Torque problem

## 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

$$\tau = \vec{\mu} B sin\vartheta$$

$$B = \frac{\mu_0}{4 \pi} \frac{2 \vec{\mu}}{z^3}$$

## The Attempt at a Solution

since its perpendicular, sin90 = 1 so:

$$\tau = \vec{\mu} B \Rightarrow \vec{\mu} = \frac{\tau}{B} = \frac{0.075}{0.5} = 0.15$$

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

Any help would be appreciated. Thank You.

i don't get it.. what is ur problem as such? as far as i see it, you have already solved it...

I inputted 37.5 and it was wrong.

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Considering that 37.5 T is 750,000 times greater than that on the surface of the earth, you might want to recalculate that value.

I see what I did wrong. I forgot to multiply it by 10^-7 so the answer should be 3.75 x 10^-6, and it is.

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