IIT JEE Physics (1978-2018: 41 Years) Topic-wise Complete Solutions

# Magnetic Swing

## Introduction

The force by a magnetic field on a current is given by $$\mathrm{d} \vec{F}=i\mathrm{d} \vec{l}\times\vec{B}$$. Three are various demonstrations based on this fact. A swing made by a copper wire is one standard demo. In our experiment, you will use a ring magnet and together with the force law, will also be able to feel the magnetic field line directions due to the ring magnet.

## Apparatus

A copper wire swing DC source, a ring magnet with a mark on North Pole.

Swing: Two aluminium rods are fixed on a plastic base. In upper portion of each rod, there is a hole. A copper wire (enameled but the ends scratched) is bent in U-shape and is suspended through the holes. This can freely swing on the holes.

## Procedure

1. Connect the DC source (keep the switch off) to the two aluminium rods using clips. Identify the direction of current (that will flow) in the horizontal part of the swing.
2. Put the ring magnet below the swing with north pole facing upward. Switch on the DC source. Which direction did the swing go?
3. Invert the ring magnet and repeat step-3. Which direction did the swing go?
4. Now shift the ring magnet in one direction and do step-3. Do it for several positions of the magnet. Locate the position for which the swing does not more when the current is switch on. Why does the swing not move in this position of magnet?
5. Shift the magnet further in the same direction further. Switch on the current. Which direction did the swing go?
6. Now keep the magnet in vertical position close to an aluminium rod. First keep the magnet outside the swing (Figure b). Pass the current and see the deflection. Now bring the magnet inside the swing but still close to the aluminium rod. Which side is the deflection?
7. Check that $$\mathrm{d} \vec{F}=i\mathrm{d} \vec{l}\times\vec{B}$$ works in every case.