A 37.9 A current flows in a long, straight wire. Find the strength of the resulting magnetic field at a distance of 45.5 cm from the w

Question

A 37.9 A current flows in a long, straight wire. Find the strength of the resulting magnetic field at a distance of 45.5 cm from
the wire.

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1 year 2021-09-03T17:56:42+00:00 1 Answers 15 views 0

$$1.67\cdot 10^{-5} T$$

Explanation:

The magnitude of the magnetic field produced by a current-carrying wire is given by the equation:

$$B=\frac{\mu_0 I}{2\pi r}$$

where:

$$\mu_0=4\pi \cdot 10^{-7} H/m$$ is the vaacuum permeability

I is the current in the wire

r is the distance from the wire

The direction of the magnetic field lines is tangential to concentric circles around the wire.

In this problem, we have:

$$I=37.9 A$$ is the current in the wire

$$r=45.5 cm = 0.455 m$$ is the distance  from the wire

Solving for B, we find the magnitude of the magnetic field:

$$B=\frac{\mu_0 I}{2\pi r}=\frac{(4\pi \cdot 10^{-7})(37.9)}{2\pi (0.455)}=1.67\cdot 10^{-5} T$$