(a) Explain giving reasons, the basic difference in converting a galvanometer into (i) a voltmeter and (ii) an ammeter. (b) Two long straight parallel conductors carrying steady currents I1 and I2 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.

(a)The attached figure shows the conversion of galvanometer into a voltmeter and ammeter respectively.When making a voltmeter, an extra series resistance of very large value is added to the galvanometer and the combination is used in parallel to the resistance under test.When making an ammeter, an extra shunt resistance of very small value is added in parallel to the galvanometer and the combination is used in series to the resistance under test (b)The attached figure shows the magnetic field and force between two parallel current carrying wires is shown in the attached figure.Consider length l of the right conductor (carrying current I2).Magnetic field at the right conductor is B1 = μoI2/2πdMagnetic force on length l of the right conductor is given by:→F = I2(→l × →B1)F2 = μoI1I2l/2πdForce acting per unit length of the conductor is:F2/l = μoI1I2/2πdBy similar arguments, it can be shown that:F1/l = μoI1I2/2πdThis is a non-contact mutual force between the two wires. It is attractive if the directions of current is same and repulsive if the directions of currents is opposite.