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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
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Answers ( )
Answer:
[tex]1.67\cdot 10^{-5} T[/tex]
Explanation:
The magnitude of the magnetic field produced by a current-carrying wire is given by the equation:
[tex]B=\frac{\mu_0 I}{2\pi r}[/tex]
where:
[tex]\mu_0=4\pi \cdot 10^{-7} H/m[/tex] 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:
[tex]I=37.9 A[/tex] is the current in the wire
[tex]r=45.5 cm = 0.455 m[/tex] is the distance from the wire
Solving for B, we find the magnitude of the magnetic field:
[tex]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[/tex]