Question

Magnetic field due to a straight current carrying conductor of finite length

Answer 1

Straight current carrying conductor of finite length –

Suppose a straight current carrying wire AB, carrying current i, lies in the plane of the paper as shown in the figure, P is a point at a perpendicular distance R from conductor, where magnetic field is to be determined.

Conductor

According to Biot-Savart’s Law, the field at point P due to current element idl = qVB→ is

dB =

μ0

4π

idlsinθ

r2

=

μ0I

4π

(

dlsinθ

r

)

i

r

Now from the figure dlsinθ = rdφ and cosφ = R/r

∴ dB =

μ0I

4π

.

cosφ

R

dφ

Hence magnetic field due to the whole conductor

B = θ1∫−θ2 = dB = θ1∫−θ2

μ0I

4πR

cosφ.dφ =

μ0I

4πR

[sinθ1 + sinθ2]

∴ dB =

μ0I

4πR

[sinθ1 + sinθ2]