## Related questions with answers

Question

(a) What is the force per meter on a lightning bolt at the equator that carries 20,000 A perpendicular to Earth’s $3.0 \times 10 ^ { - 5 } \mathrm { T }$ field? (b) What is the direction of the force if the current is straight up and Earth’s field direction is due north, parallel to the ground?

Solution

VerifiedStep 1

1 of 3a) The magnitude of the force on a wire of length $l$ carrying a current $I$ and making an angle $\theta$ with a magnetic field of density $B$ is given by:

```
$\vert \vec{F} \vert = B \, I \, l \, \sin{(\theta)}$
```

Rearranging to get the force per unit length:

```
$\dfrac{\vert \vec{F} \vert}{l} = B \, I \, \sin{(\theta)}$
Therefore,
```

$\begin{align*} \dfrac{\vert \vec{F} \vert}{l} &= (3\times10^{-5}) \times (2\times 10^4) \times \sin(\dfrac{\pi}{2})\\ &=0.6 \, \mathrm{N.m^{-1}} \end{align*}$

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