## Long Straight Current Carrying Wire

**Long Straight Current Carrying Wire** Assignment Help | **Long Straight Current Carrying Wire** Homework Help

# Magnetic Field for a Long Straight Current Carrying Wire

Consider a wire of length L, carrying a current I. Let P be a point at a distance x from the wire. Let O be the origin of coordinate system, with y-axis along the current and x-axis towards the field point P. Let dy be a small element at a distance y from O. Let j be unit vector along y-axis.The magnetic potential at P due to the element dy is

dA = μ

_{0}I/4π dy/r directed along the current.

= μ

_{0}I/4π dy/√(x

^{2}+ y

^{2}) j … (1)

The magnetic potential at P due to the whole wire is

A = j μ

_{0}I/4π ʃ dy/√(x

^{2}+ y

^{2})

= j μ

_{0}I/4π [log {y + √(x

^{2}+ y

^{2})}] +L/2

= j μ

_{0}I/4π log {L/2 + √(L2/4 + x

^{2}) / - L/2 + √(L2/4 + x

^{2})}

= j μ

_{0}I/4π log [L/2{1 + (1 + 4x

^{2}/L

^{2})1/2} / L/2 {-1 + (1 + 4x

^{2}/L

^{2})1/2}] … (2)

If the wire is infinitely long, then x2/L2 << 1. Hence Eq. (2) reduces to

A = j μ

_{0}I/4π log [1 + (1 + 4x

^{2}/L2) } / -1 + (1 + 4x

^{2}/L2)]

= j μ

_{0}I/4π log [(2 + 2x

^{2}/L2) / (2x

^{2}/L2)]

= j μ

_{0}I/4π log (1 + L2/x

^{2})

= j μ

_{0}I/4π log (L/x) 2 (. : L

^{2}/x

^{2}>> 1)

. : A = (μ

_{0}/4π) 2I log (L/x) j … (3)

This is required expression for A.

. : B = ∇ x A = ∇ x (μ0/4π 2l log (log (L/x) j)

= μ

_{0}/4π 2I ∂/∂x [log (l/x)] k

. : B = μ

_{0}/4π (2I/x) k

The magnetic field has magnitude μ

_{0}/4π (2l/x) and is directed perpendicular to plane of paper downwards.

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