A horizontal copper rod of mass 12 g and length 0.2 m lies east to west in a horizontal field of…

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Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A horizontal copper rod of mass $12\ \text{g}$ and length $0.2\ \text{m}$ lies east to west in a horizontal field of $0.15\ \text{T}$ that points north. Take $g = 9.8\ \text{m/s}^2$ .

What current makes the magnetic force on the rod equal to its weight?
Which way must the current flow for the force to point up?

Need a reminder? The key ideaShow

A wire with a current in a magnetic field feels a force $F = BIL\sin\theta$ , where $\theta$ is the angle between the wire and the field. It is largest ( $F = BIL$ ) at right angles, and zero along the field. Fleming's left-hand rule: forefinger for the field, middle finger for the current, thumb for the force.

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

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