## Moving charges and magnetism

# Magnetic Field on the Axis of a Circular Current Loop, Ampere’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 i
_{1}, i_{2}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 = B
_{1}+ B_{2}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}

### Ampere's law View the Topic in this video From 01:44 To 41:34

### 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 μ_{0} 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}