 # Magnetic Field on the Axis of a Circular Current Loop, Ampereâ€™s Circuital Law

• The magnetic field at the centre of the coil is \tt B = \frac{\mu_{0} ni}{2r}
• The magnetic field at the centre when conductor is bent in the form of a coil. \tt B = \frac{\mu_{0} ni}{4 \pi r} (\theta) (θ in radius)
• If the wire is in the form of semi-circle \tt B = \frac{\mu_{0} i}{4 \pi r} (\pi) = \frac{\mu_{0}i}{4 r}
• For comparing magnetic fields \tt \frac{B_{1}}{B_{2}} = \left(\frac{n_{1}}{n_{2}}\right)^{2} = \left(\frac{r_{1}}{r_{2}}\right)^{2}
• If two circular coils are connected in series. \tt \frac{B_{1}}{B_{2}} = \left(\frac{n_{1}}{n_{2}}\right) \left(\frac{r_{1}}{r_{2}}\right)
• If two coils are connected in parallel. \tt \frac{B_{1}}{B_{2}} = \left(\frac{r_{2}}{r_{1}}\right)^{2}
• If two concentric circular loops carrying currents i1, i2 and their planes are inclined at an angle θ. The resultant magnetic induction at the common centre is \tt BR = \sqrt{B_{1}^{2} + B_{2}^{2} + 2 B_{1}B_{2} \cos \theta }
• If θ = 0 ⇒ B = B1 + B2 and If θ = 90° then \tt B = \sqrt{B_{1}^{2} + B_{2}^{2}}
• Null point in the following situation \tt x = \frac{d}{\left(\frac{i_{1}}{i_{2}}\right)^{1/3} \pm 1} • Magnetic field due to straight conductor of finite length at a perpendicular distance “r” is \tt B = \frac{\mu_{0} i}{4 \pi r} (\sin \theta_{1} + \sin \theta_{2}) • Magnetic induction at a point due to a straight conductor of infinite length at ‘r’ \tt B = \frac{\mu_{0} i}{2 \pi r}
• If the point is ‘φ’ along the length of the conductor at that point B = 0
• “B” at a perpendicular distance r for infinite wire is \tt B = \frac{\mu_{0} i}{4 \pi r}
• ‘B’ at a perpendicular distance r for a finite conductor is \tt B = \frac{\mu_{0} i}{4 \pi r} \frac{l}{\sqrt{l^{2} + r^{2}}}
• Ampere’s circuital law used for calculating the magnetic field under highly symmetrical conditions and is given by \tt \oint B . dl = \mu_{0} i
• Force acting on a magnetic pole (m) kept at a distance “r” from a infinite current carrying conductor is \tt F = \frac{\mu_{0} i}{2 \pi r} m
• If magnetic pole ‘m’ is n times moved around the conductor were done \tt \oint \overline{f} . \overline{d} s = \frac{\mu_{0} i m}{2 \pi r} . 2 \pi r = \mu_{0} i m n
• The magnetic field induction inside the current carrying vary long solid cylinder at a distance ‘r’ is \tt B = \frac{\mu_{0} i r}{2 \pi R^{2}} (R = Radius of the conductor) (r < R) \tt B = \frac{\mu_{0} i }{2 \pi R} (r > R)
• Magnetic induction at the centre of current carrying wire bent in the form of a square “l” is \tt B = 8\sqrt{2} \frac{\mu_{0}}{4 \pi} \left(\frac{i}{l}\right) • Magnetic induction at the centre of current carrying wire bent in the form of a triangle of side “l” is \tt B = 18 \frac{\mu_{0} i}{4 \pi l} • Magnetic induction at the centre of current carrying wire bent is the form of hexagon of side ‘l’ is \tt B = 4\sqrt{3} \frac{\mu_{0}}{4 \pi} \left(\frac{i}{a}\right) • \tt B_{P} = \frac{\mu_{0}}{2 \pi} \left[\frac{i_{1}}{r_{1}} - \frac{i_{2}}{(r - r_{1})}\right]
\tt B_{Q} = \frac{\mu_{0}}{2 \pi} \left[\frac{i_{1}}{r_{2}} + \frac{i_{2}}{(r + r_{2})}\right]
• \tt B_{P} = \frac{\mu_{0}}{2 \pi} \left[\frac{i_{1}}{r_{1}} + \frac{i_{2}}{(r - r_{1})}\right]

\tt B_{Q} = \frac{\mu_{0}}{2 \pi} \left[\frac{i_{1}}{r_{2}} + \frac{i_{2}}{(r + r_{2})}\right]

• Null point position \tt S_{1} = \frac{r}{\frac{i_{2}}{i_{1}}+1} • Null point position \tt S_{2} = \frac{r}{\frac{i_{2}}{i_{1}}-1} ### View the Topic in this video From 41:14 To 55:45

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1. Ampere's Circuital Law
It states that the line integral of magnetic field around any closed path in vacuum is equal to μ0 times the total current passing the closed path.
\oint \overrightarrow{B}.\overrightarrow{dl} = \mu_{0}l

2. Magnetic field at a point outside the wire i.e. (r > a) is
B = \frac{\mu_{0}I}{2 \pi r}