Question
  1. Explain, giving reasons, the basic difference in converting a galvanometer into $(i)$ a voltmeter and $(ii)$ an ammeter.
  2. Two long straight parallel conductors carrying steady currents $I_1$ and $I_2$ are separated by a distance $'d\ '$. Explain briefly, with the help of a suitable diagram, how the magnetic field due to one conductor acts on the other. Hence deduce the expression for the force acting between the two conductors. Mention the nature of this force.

Answer

  1. A galvanometer can be converted into a voltmeter by connecting a high resistance in series with its coil.
  2. A galvanometer can be converted into an ammeter by connecting a low resistance in parallel to its coil.
Alternate Answer



The magnetic field, due to wire $– 1,$ at any point on the wire $– 2,$ is directed normal to the direction of current flowin wire $– 2.$
Expression for force
$B_{21} =$ Magnetic field atwire $– 2,$ due to a current $I_1$ in
wire $ - 1 = \frac{\mu_{o}\text{I}_{1}}{2\pi\text{r}}$
Force upon conductor carrying current due to magnetic field
$\overrightarrow{\text{F}} = \text{I}(\overrightarrow{\ell}\times\overrightarrow{\text{B}})$
$\therefore\text{ Force, F}_{21}\text{ on a length l of wire}\ – 2 = \text{I}_{2} . l.\frac{\mu_{o}\text{I}_{1}}{2\pi\text{r}}$
$ =\frac{\mu_{o}\text{I}_{1}\text{I}_{2}}{2\pi\text{r}}l$
$\text{Similarly,} = \text{F}_{12} = \frac{\mu_{0}\text{I}_{1}\text{I}_{2}}{2\pi\text{r}}l$
Nature
The force is repulsive for currents in opposite direction and attractive when currents flow in the same direction.

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