Q. 225.0( 2 Votes )

Describe the work

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 SP 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.


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 S1 by r1 and the number of turns per unit length by n1. The same physical quantities for the outer solenoid S2 are r2 and n2, respectively. Let N1 and N2 be the total number of turns of coils S2 and S2, respectively.

Derivation: -

When a current I2 starts to flow in S2, it in turn sets up a magnetic flux through S1. Let us denote it by F1. The corresponding flux linkage with solenoid S1 is

N1 = M12I2 ….(a)

The mutual inductance of solenoid S1 with respect to solenoid S2 is known as M12. The magnetic field due to the current I2 in S2 is,

B = μ0n2I2

The resulting flux linkage with coil S1 is,

N1 = (n1l) (r12) (n2I2)

N1 = n1n2 r12l I2

where the total number of turns in solenoid S1 is denoted n1l. Thus, from Eq. (a), we can write,

Conclusion: -

The mutual inductance is

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses

Describe the workPhysics - Board Papers

A. Find the valuePhysics - Board Papers

A wire in the forPhysics - Exemplar

(a) Define the tePhysics - Board Papers

Define ‘mutual inPhysics - Exemplar

Consider the enerHC Verma - Concepts of Physics Part 2

A <span lang="EN-Physics - Exemplar