Answer
Verified
105.6k+ views
Hint: To solve this question we should know what a tangent galvanometer is. After that, we can understand what the reduction factor for a tangent galvanometer is and the conditions associated with it. Using this we can easily solve the required solution for the question.
Formulae used:
$I = K\tan \theta $
Where $I$ is the current supplied, $K$ is the reduction factor and $\theta $ is the angle of deflection shown by the galvanometer.
$\tan \theta = \dfrac{{{B_V}}}{{{B_H}}}$
Here $\theta $ is the angle of deflection shown by the galvanometer, ${B_V}$ is the vertical component of magnetic field and ${B_H}$ is the horizontal component of magnetic field.
Complete step by step solution:
We know that,
$I = K\tan \theta $
Where $I$ is the current supplied, $K$ is the reduction factor of the galvanometer, and $\theta $ is the angle of deflection shown by the galvanometer.
To obtain the reduction factor, the tangent galvanometer should be in balanced condition i.e.
$ \Rightarrow {B_V} = {B_H}$
Let this be equation 1.
Here ${B_V}$ is the vertical component of the magnetic field of earth and ${B_H}$ is the horizontal component of magnetic field of the earth.
We know that
$ \Rightarrow \tan \theta = \dfrac{{{B_V}}}{{{B_H}}}$
Substituting the of equation 1 here, we get,
$ \Rightarrow \tan \theta = \dfrac{{{B_V}}}{{{B_V}}} = 1$
$ \Rightarrow \tan \theta = \tan 45^\circ $
$ \Rightarrow \theta = 45^\circ $
So we get the angle of deflection as $45^\circ $. So the reduction factor of a tangent galvanometer may be defined as the current passing through it to produce a deflection of $45^\circ $.
Hence the required solution will be option (B).
Note: A tangent galvanometer is a measuring instrument used for the measurement of electric current. It works on the principle of the tangent law of magnetism. It uses a copper coil wound on a nonmagnetic frame. It has a compass needle placed horizontally to the surface. It compares the magnetic field generated by the unknown current to the magnetic field of the Earth.
Formulae used:
$I = K\tan \theta $
Where $I$ is the current supplied, $K$ is the reduction factor and $\theta $ is the angle of deflection shown by the galvanometer.
$\tan \theta = \dfrac{{{B_V}}}{{{B_H}}}$
Here $\theta $ is the angle of deflection shown by the galvanometer, ${B_V}$ is the vertical component of magnetic field and ${B_H}$ is the horizontal component of magnetic field.
Complete step by step solution:
We know that,
$I = K\tan \theta $
Where $I$ is the current supplied, $K$ is the reduction factor of the galvanometer, and $\theta $ is the angle of deflection shown by the galvanometer.
To obtain the reduction factor, the tangent galvanometer should be in balanced condition i.e.
$ \Rightarrow {B_V} = {B_H}$
Let this be equation 1.
Here ${B_V}$ is the vertical component of the magnetic field of earth and ${B_H}$ is the horizontal component of magnetic field of the earth.
We know that
$ \Rightarrow \tan \theta = \dfrac{{{B_V}}}{{{B_H}}}$
Substituting the of equation 1 here, we get,
$ \Rightarrow \tan \theta = \dfrac{{{B_V}}}{{{B_V}}} = 1$
$ \Rightarrow \tan \theta = \tan 45^\circ $
$ \Rightarrow \theta = 45^\circ $
So we get the angle of deflection as $45^\circ $. So the reduction factor of a tangent galvanometer may be defined as the current passing through it to produce a deflection of $45^\circ $.
Hence the required solution will be option (B).
Note: A tangent galvanometer is a measuring instrument used for the measurement of electric current. It works on the principle of the tangent law of magnetism. It uses a copper coil wound on a nonmagnetic frame. It has a compass needle placed horizontally to the surface. It compares the magnetic field generated by the unknown current to the magnetic field of the Earth.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
Write an article on the need and importance of sports class 10 english JEE_Main
Name the scale on which the destructive energy of an class 11 physics JEE_Main
Choose the exact meaning of the given idiomphrase The class 9 english JEE_Main
Choose the one which best expresses the meaning of class 9 english JEE_Main