## Related questions with answers

At a location in the Philippines, the Earth's magnetic field has a value of $39.0 \mu \mathrm{T}$ and is horizontal and due north. The net field is zero $8.13 \mathrm{~cm}$ above a long, straight, horizontal wire that carries a steady current. (a) Calculate the current and (b) find its direction.

Solution

Verified$\text{\underline{Part a}}$.

We know that the magnitude of the magnetic field produced by a wire carrying current $I$ at a distance $d$ away is given by

$B=\frac{\mu_0 I}{2\pi d}.$

From this, the current needed to create a certain field will be given as

$I=\frac{2\pi dB}{\mu_0}.$

The field that needs to be created has just the same magnitude as the one due to the Earth. Thus, we can substitute numerically, and find the current to be

$I=\frac{2\pi \cdot 0.0813\cdot 39.0\cdot 10^{-6}}{4\pi \cdot 10^{-7}}=\boxed{15.9~\mathrm{A}}.$

$\text{\underline{Part b}}$.

The Earth's magnetic field is horizontal due north. For the net field to be zero, the field due to the current will have to be horizontal due south.

It is clear, since the field that a current produces lies in the cross-section plane, that the direction of the current will be horizontal, in the east-west line. Now, let's find due where. Since the field should be zero somewhere above the wire, using the right hand grip rule to obtain a field pointing due south above the wire, the direction of the current will be $\text{\underline{due east}}$.

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