Answer :

a) The moving coil galvanometer works on the principle that a current carrying loop experiences a torque when placed in external magnetic field. The Moving coil galvanometer consists of a coil having large number of turns, which is allowed to rotate about a fixed axis, in a uniform radial magnetic field. Radial magnetic field is used as it ensures that maximum torque is applied to the coil at all angles and there is a cylindrical soft iron core which not only makes the field radial but also increases the strength of the magnetic field. When a current flows through the coil, a torque acts on it.

This torque is given by.

N is the number of turns, I is current, A is the area of c.s., B is the strength of the magnetic field.

The magnetic torque NIAB tends to rotate the coil. A spring S_{P} provides a counter torque k that balances the magnetic torque NIAB, resulting in a steady angular deflection . In equilibrium,

where k is the torsional constant of the spring, i.e. the restoring torque per unit twist. The deflection is indicated on the scale by a pointer attached to the spring. We have.

The term NAB/k is constant for a given galvanometer and is referred to as current sensitivity, i.e.,

No a galvanometer cannot be used to measure current as such because it is highly sensitive to even slight fluctuation in current. If such a large values of current can destroy the galvanometer.

OR

i) When an EMF is induced in a single isolated coil due to change of flux through the coil by the virtue of the varying current through the same coil is called Self Induction. In this case, flux linkage through a coil of N turns is proportional to the current through the coil and is expressed as,

Where, L is called self-inductance of the coil. The SI unit of self-inductance is Henry (H).

ii) Given: -

Two long co-axial solenoids each of length l. The radius of the inner solenoid S_{1} by r_{1} and the number of turns per unit length by n_{1}. The same physical quantities for the outer solenoid S_{2} are r_{2} and n_{2}, respectively. Let N_{1} and N_{2} be the total number of turns of coils S_{2} and S_{2}, respectively.

Derivation: -

When a current I_{2} starts to flow in S_{2}, it in turn sets up a magnetic flux through S_{1}. Let us denote it by F_{1}. The corresponding flux linkage with solenoid S_{1} is

N_{1} = M_{12}I_{2} ….(a)

The mutual inductance of solenoid S_{1} with respect to solenoid S_{2} is known as M_{12}. The magnetic field due to the current I_{2} in S_{2} is,

B = μ_{0}n_{2}I_{2}

The resulting flux linkage with coil S_{1} is,

N_{1} = (n_{1}l) (r_{1}^{2}) (n_{2}I_{2})

N_{1} = n_{1}n_{2} r_{1}^{2}l I_{2}

where the total number of turns in solenoid S_{1} is denoted n_{1}l. Thus, from Eq. (a), we can write,

Conclusion: -

The mutual inductance is

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