Search Coil

Search coil

A search coil is a solenoid of small diameter having few (about 50) turns of insulated copper wire and is suitable for measuring strong magnetic field.

Suppose, we need to find the magnetic field between two poles pieces of an electromagnet. The search coil is placed in gap between the pole pieces and is connected with a ballistic galvanometer through high resistance ‘R’ to control the thrown in B.G and secondary coil ‘SS’ of standard solenoid as in fig.

We can change the magnetic flux in two ways:

1) Put on the magnetic field of the electromagnetic and search coil is either suddenly inserted or withdrawn between poles of magnet.

2) Hold the search coil in between poles and magnetic field is suddenly put on.

Now, as the flux linked with search coil is changed, so, induced emf is observed which gives deflection in B.G.

Suppose the deflection be \(\begin{align*}\theta\end{align*}\)

No. of turns in search coil= N

Area of each turn of coil= A

Magnetic field= B

And, we have flux link with coil \(\begin{align*}(\phi )=NBA\end{align*}\)

\begin{align*}d\phi =NAB\end{align*}

Induced Emf (E) \(\begin{align*}=\frac{d\phi }{dt}=\frac{NAB}{dt}\end{align*}\)

And E= IR

\begin{align*}I=\frac{NAB}{Rdt}-----(i)
\end{align*}

And charge flowing in time ‘dt’ q= Idt

\begin{align*}q=\frac{NAB}{R}-----(ii)\end{align*}

But in case of B.G charge flow \(\begin{align*}=k\theta (1+\frac{\lambda }{2})\end{align*}\)

Where, K= ballistic constant

\begin{align*}\frac{NAB}{R}=k\theta (1+\frac{\lambda }{2})-----(iii)\end{align*}

For \(\begin{align*}\theta\end{align*}\)is deflection angle and \(\begin{align*}\lambda\end{align*}\) logarithm decrement in angle------------ (iii)

Now, to determine K we connect the galvanometer in series with an ammeter and rheostat which is further connected to primary of standard solenoid. A known current in primary circuit is passed. Now, it produces field inside solenoid

\begin{align*}B_{p} =\mu n_{p}I\end{align*}

This cause magnetic flux to pass through the secondary and flux linked with secondary coil.

\begin{align*}\oint =B_{p}N_{s}A_{s}=\mu n_{p}IN_{s}A_{s}\end{align*}

For np= no. of turns per unit length in primary coil.

Ns= turns in secondary coil

As= Area of secondary coil

I= current in primary coil

Now, reverse the direction of ‘I’ so, total flux change \(\begin{align*}=\mu n_{p}IN_{s}A_{s}-(-\mu n_{p}IN_{s}A_{s})\end{align*}\)

\begin{align*}\oint _{total}=2\mu n_{p}IN_{s}A_{s}\end{align*}

Charge due to this flux \(\begin{align*}=\frac{2\mu n_{p}IN_{s}A_{s}}{R}----(iv)\end{align*}\)

Let \(\begin{align*}\theta _{1}\end{align*}\) be the deflection in B.G then,

\begin{align*}\frac{2\mu n_{p}IN_{s}A_{s}}{R}=k\theta _{1}(1+\frac{\lambda }{2})-----(v)\end{align*}

Dividing (iii) by (v) then,

\begin{align*}B=\frac{2\mu n_{p}IN_{s}A_{s}\theta }{NA\theta _{1}}-----(*)\end{align*}

In case rather than reversing the direction of ‘I’ if the current is suddenly stopped or started, flux linked will be \(\begin{align*}\mu n_{p}IN_{s}A_{s}\end{align*}\) throw \(\begin{align*}\theta _{2}\end{align*}\) in B.G will now be half of \(\begin{align*}\theta _{1}\end{align*}\) then

\begin{align*}B=\frac{\mu n_{p}IN_{s}A_{s}\theta }{NA\theta _{2}}-----(**)\end{align*}

Hence, (*) and (**) are the way to calculate field.

References

Adhikari, Pitri Bhakta. A Textbook of Physics Volume-I. Kathmandu: Sukunda Pustak Bhawan, 2015.

Feynman, Richard P. The Feynman Lectures on Physics Volume 1. Noida: Dorling Kindersley (India) Pvt. Ltd., 2014.

Mathur, D S. Mechanics. New Delhi: S. Chand & Company Pvt. Ltd., 2015.

Young, Hugh D, Roger A Freedman and A Lewis Ford. University Physics. Noida: Dorling Kindersley (India) Pvt. Ltd., 2014