# Magnetic field at the center of an arc of current – problems and solutions

1. Based on the figure below, if the radius of the curvature of the wire is 50 cm, determine the magnitude of the magnetic field at the center of curvature (at point 0, see figure below). (µ_{o} = 4π.10^{-7} Wb.A^{-1} m^{-1})

__Known :__

Radius (r) = 50 cm = 0.5 m

Electric current (I) = 1.5 Ampere

The vacuum permeability (µ_{o}) = 4π.10^{-7 }Wb.A^{-1} m^{-1}

__Wanted:__ The magnitude of the magnetic field

__Solution :__

360^{o} = 1 *circumference* of a circle. 120^{o} / 360^{o} = 1/3 then 120^{o} = 1/3 x *circumference* of a circle.

__The equation of the magnetic field at the center of the coil with a number of loops :__

*B = the magnitude of the magnetic field, N = number of loops, I = electric current, r = radius of curvature*

In the above problem, there is only one loop so that N is eliminated from the equation. The wire coil on the above problem is not 1 circle but 1/3 circle :

__The magnitude of the magnetic field at the center of curvature :__

2. Based on the figure below, the electric current flows in the wire is 6-A and radius of curvature is R = 3π cm, determine the magnitude of the magnetic field at point P.

__Known :__

Radius of curvature (r) = 3π cm = (3π/100) m

= 3π/10^{2} m = 3π.10^{-2 }m

Electric current (I) = 6 A

The vacuum permeability (µ_{o}) = 4π.10^{-7 }Wb.A^{-1} m^{-1}

__Wanted :__ The magnitude of the magnetic field

__Solution :__

360^{o} = 1 *circumference* of a circle. 45^{o} / 360^{o} = 1/8 then 45

^{o}= 1/8 x

*circumference*of a circle.

__The magnitude of the magnetic field at the center of curvature :__

3. Electric current flows in wire = 9-A, the radius of curvature (R) = 2π cm and µ_{o} = 4π.10^{-7} Wb.A^{-1}.m^{-1}, determine the magnitude of the magnetic field at point P.

__Known :__

Radius of curvature (r) = 2π cm = (2π/100) m

= 2π/10^{2} m = 2π.10^{-2 }m

Electric current (I) = 9 A

The vacuum permeability (µ_{o}) = 4π.10^{-7 }Wb.A^{-1} m^{-1}

__Wanted :__ The magnitude of the magnetic field at point P

__Solution :__

360^{o} – 120^{o} = 240^{o}. 240^{o} / 360^{o} = 2/3 then 240^{o} = 2/3 x *circumference* of a circle.

__The magnitude of the magnetic field at the center of curvature :__