Question

coil has 30 turns and carries current. The cross sectional area of the coil is 7.5cm. At a point located on the axis of the coil at a distance 1.5m, the magnetic field os 0.2mG, what is current in the coil?

Answer 1

Except along the axis ,the magnetic field of a circular coil cannot be expressed in closed form .Along the coil axis ,if the origin of the coordinates is taken at the centre of the coil and if the z-axis is taken along the coil axis ,the magnitude of the magnetic field B,which points in the z-direction ,is given by,

B=₀N*I*a² / 2(a² +z²)^3/2

where N is no.of turns of the coil

          I is the current

          a is the radius of the coil

          z is axial distance

          ₀= 4*pie*10^-7 vaccum permeability

in the problem given values are

N=30 turns

z=1.5 m

area=pie *r² =7.5cm

radius a= 0.015454886 m

          a=0.015m

B=0.2mG=2*10^-8 Tesla

sustituting the values we get,

2*10^-8 =4*3.14* 10^-7*30* (0.015)² *I / 2*((0.015)² + (1.5)²)^3/2

we get current I =26.87898 amp