## Flat Circular Magnetic Shell

**Flat Circular Magnetic Shell** Assignment Help | **Flat Circular Magnetic Shell** Homework Help

# Magnetic Potential and Field at a Point on the Axis of a Flat Circular Magnetic Shell

Consider a flat circular magnetic shell ABC of uniform strength Φ. Let O be its centre and a, its radius. P is a point on the axis at a distance x from O. Let ∟APO = θ.If the shell subtends a solid angle ω at P, the potential at P is

V = μ

_{0}/4π Φω

ω is determined as follows:

Let ADEC be the side view of the shell having width ds.

Talking P as centre, draw a sphere of radius PA = r. The circumference of the shell then lies on the surface of the sphere. The solid angle subtended by the shell at P is given by

ω = Area of the cap ABC/r

^{2}

The area of the rim of the shell = (2πa)ds

The area of cap ABC can be calculated by integrating this area between the limits 0 and θ.

Area of cap ABC = ∫

^{θ}

_{0}2π a ds

a = r sin θ and ds = r dθ

. : area of the cap = ∫

^{θ}

_{0}2π r

^{2}sin θ dθ = 2πr

^{2}(1-cosθ)

Solid angle subtended by the cap ABC at P is

ω = 2πr

^{2}(1-cos θ)/r

^{2}= 2π (1 – cos θ)

= 2π (1 – x/r) = 2π (1 – x/ (a

^{2}+ x

^{2})

^{1/2})

The potential at P is given by

V = μ

_{0}/4π Φω

V = μ

_{0}/4π [2πΦ (1 – x/(a

^{2}+ x

^{2})]

If x = 0, i.e., if the point is very close the shell, V = μ0Φ/2

Magnetic Induction field (B)

Magnetic induction B = -dV/dx

. : B = -d/dx [μ

_{0}Φ/2{1 – x/ (a2 + x2)

^{1/2}}] = μ0Φa

^{2}/2(a

^{2}+ x

^{2})

^{3/2 }

If x = 0, then B = μ0Φ/2a

For more help in

**Magnetic Potential and Field at a Point on the Axis of a Flat Circular Magnetic Shell**please click the button below to submit your homework assignment.