Question

what is the flux density at a point 3 cm from the long straight wire when there is a current of 25 A in the wire

Answer 1

Answer:

4.69*10^-5.

Explanation:

We know that the magnetic field of a wire is given by the formulae of B= μi/2πr where i is the current flowing through the wire and r is the radius of the wire.

B= 4π*10^-7*25/2π*3 which on solving we will get 16.6*10^-7 T and the flux flowing will be B.A or 16.6*10^-7*π(3)^2 or 4.69*10^-5.

Here μ is the permeability of free space which is having the value of 4π*10^-7 T m/A.

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Answer 2

answer : 1.667 × 10^-4 T

explanation : first understand what is the meaning of magnetic flux density ?

magnetic flux density is the number of lines of force that passing through a point on the specific surface.

i.e., Magnetic flux density = Φ/A

we know, Φ = BA

so, magnetic flux = BA/A = B ( magnetic field.)

magnetic field at a point of a long straight wire is given as , $\frac{\mu_0I}{2\pi r}$

given, I = 25 A and r = 3cm = 0.03m

= 4π × 10^-7 × 25/2π × (0.03)

= (50/3) × 10^-5 T

= 1.667 × 10^-4 T

hence, magnetic flux density is 1.667 × 10^-4 T .