 # The Solenoid and the Toroid, Force between Two Parallel Currents, the Ampere

• Magnetic induction at a point on the axis of solenoid \tt B = \frac{\mu_{0} n i}{2} (\sin \alpha + \sin \beta) • "B" near one end \tt B = \frac{\mu_{0} ni}{2}
• Magnetic induction at the centre of finite solenoid \tt B = \frac{\mu_{0}}{4 \pi} (4 \pi n i) \sin \alpha\tt and  \sin \alpha = \frac{L}{\sqrt{L^{2} + 4 R^{2}}}
• Magnetic induction in toroid \tt B = \frac{\mu_{0}}{2 \pi} . \frac{Ni}{r} (L = 2 \pi r)
• B = \frac{\mu_{0} Ni_{0}}{L}
• Magnetic field inside the core and the field outside the toroid is zero i.e. BQ = 0 and BR = 0.
• Force between two current carrying parallel wires is \tt \frac{F}{l} = \frac{\mu_{0}}{2 \pi} \frac{i_{1} i_{2}}{r}
• If currents in the two wires are in same direction then the force of attraction taken place between them.
• If currents in the two wires are in opposite direction then the force of repulsion taken place between them.
• Force on the loop is \tt F = \frac{\mu_{0} i_{1} i_{2} l}{2 \pi} \left[\frac{1}{x} - \frac{1}{y}\right] • Force on the hanging wire to remain stationary \tt F = mg = \frac{\mu_{0} i_{1} i_{2} l}{2 \pi r}
\frac{m}{l} = \frac{\mu_{0}i_{1}i_{2}}{2\pi r g} ### View the Topic in this video From 01;02 To 38;29

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Force Between Two Parallel Current Carrying Conductors
f = \frac{\mu_{0}}{4 \pi} \frac{2I_{1}I_{2}}{r}