Question

using bio savart paw,find expression for the magnetic field at the center of a current carrying arc of radius r and carrying current i.​

Answer 1

Answer:

The magnitude of magnetic field due to a current element according to Biot-Savart law is given by :

đB=μIđlsinθ/4πr^2

But, θ=90° and sin90°=1. Also, r = R = Radius of circular wire.

⇒đB = μIđl/4πR^2

All parameters except "đl" in the right hand expression of the equation are constants and as such they can be taken out of the integral.

B = ∫đB = (μI/4πR^2) ∫đl

The integration of dl over the complete circle is equal to its perimeter i.e. 2πR.

⇒B= (μI/4πR^2) X 2πR = μI/2R

If the wire is a coil having N circular turns, then magnetic filed at the center of coil is reinforced N times :

B = μNI/2R

Answer 2

Answer:

I already write them!!!!!!