Question A current of 1.41 A in a long, straight wire produces a magnetic field of 5.61 uT at a certain distance from the wire. Find this distance.

Answer: 0.050 m Explanation: The strength of the magnetic field produced by a current-carrying wire is given by [tex]B=\frac{\mu_0 I}{2\pi r}[/tex] where [tex]\mu_0=4\pi \cdot 10^{-7} H/m[/tex] is the vacuum permeability I is the current in the wire r is the distance from the wire And the magnetic field around the wire forms concentric circles, and it is tangential to the circles. In this problem, we have: [tex]I=1.41 A[/tex] (current in the wire) [tex]B=5.61\mu T=5.61\cdot 10^{-6} T[/tex] (strength of magnetic field) Solving for r, we find the distance from the wire: [tex]r=\frac{\mu_0 I}{2\pi B}=\frac{(4\pi \cdot 10^{-7})(1.41)}{2\pi (5.61\cdot 10^{-6})}=0.050 m[/tex] Log in to Reply

Answer:0.050 m

Explanation:The strength of the magnetic field produced by a current-carrying wire is given by

[tex]B=\frac{\mu_0 I}{2\pi r}[/tex]

where

[tex]\mu_0=4\pi \cdot 10^{-7} H/m[/tex] is the vacuum permeability

I is the current in the wire

r is the distance from the wire

And the magnetic field around the wire forms concentric circles, and it is tangential to the circles.

In this problem, we have:

[tex]I=1.41 A[/tex] (current in the wire)

[tex]B=5.61\mu T=5.61\cdot 10^{-6} T[/tex] (strength of magnetic field)

Solving for r, we find the distance from the wire:

[tex]r=\frac{\mu_0 I}{2\pi B}=\frac{(4\pi \cdot 10^{-7})(1.41)}{2\pi (5.61\cdot 10^{-6})}=0.050 m[/tex]