A wire carries 7.5 A. It comes in along a straight line that points at a point O…

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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 wire carries $7.5$ A. It comes in along a straight line that points at a point $O$ . Then it goes round an arc of radius $12$ cm and angle $240^\circ$ about $O$ . It leaves along another straight line that points away from $O$ . What is the field at $O$ ? What would a full circle of the same radius and current give at its centre?

Need a reminder? The key ideaShow

A short piece of current makes $dB = \dfrac{\mu_0}{4\pi}\dfrac{I\,dl\sin\theta}{r^2}$ , with $\mu_0 = 4\pi \times 10^{-7}\ \text{T m/A}$ . A very long wire gives $B = \dfrac{\mu_0 I}{2\pi d}$ . An arc of angle $\theta$ gives $B = \dfrac{\mu_0 I\theta}{4\pi R}$ at its centre, and $N$ full turns give $B = \dfrac{\mu_0 N I}{2R}$ .

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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