Question

The magnitude of the magnetic field that a long and extremely thin current-carrying wire produces at a distance of 3.0 µm from the center of the wire is 2.0 × 10-3 T. How much current is flowing through the wire?

Answer 1

Answer: 0.03002A

Explanation: The formulae that relates the magnetic field strength B at a point (r) away from the center of a conductor carrying a current of value (I) is given below as

B = Uo×I/2πr

From our question, B =2.0×10^-3 T, r = 3.0×10^-6m

I =?, Uo = permeability of free space = 1.256×10^-6 mkg/s²A².

By substituting the parameters, we have that

2×10^-3 = 1.256×10^-6 × I/2π(3.0×10^-6)

2×10^-3 × 2π(3.0×10^-6) = 1.256×10^-6 × I

3.77×10^-8 = 1.256×10^-6 × I

I = 3.77×10^-8/ 1.256×10^-6

I = 3.002×10^-2 = 0.03002A

Answer 2

Answer:

Current (I) = 3 x 10^-2 A

Explanation:

As we know, $B = 4\pi 10^-7 *l/ 2\pi r$

By putting up the values needed from the data...

Current (I) = 2 x 3.14 x (3.0 x 10^-6) (2.0 x 10^-3) / 4 x 3.14 x 10^-7 = 3 x 10^-2 A