(A) The principle of a tangent galvanometer is based on the tangent law, which states that $I = K \tan \theta$, where $I$ is the current, $\theta$ is

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(A) The principle of a tangent galvanometer is based on the tangent law, which states that $I = K \tan \theta$, where $I$ is the current, $\theta$ is the deflection, and $K$ is the reduction factor.From this equation, $K = I / \tan \theta$.Since $\tan \theta$ is a dimensionless quantity, the unit of the reduction factor $K$ is the same as the unit of current $I$.Therefore, the unit of the reduction factor of a tangent galvanometer is Ampere $(A)$.

Class 12 Physics Magnetism and Matter Magnetic Equipment's (Tangent galvanometer, Vibration magnetometer and Neutral point)

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The unit of the reduction factor of a tangent galvanometer is

A
Ampere
B
Gauss
C
Radian
D
None of these

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Magnetism and Matter

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Magnetic Equipment's (Tangent galvanometer, Vibration magnetometer and Neutral point)

Two tangent galvanometer coils of the same radius are connected in series. The current flowing produces deflections of $60^o$ and $45^o$. The ratio of the number of turns in the coils is:

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$A$ bar magnet is placed on a table along the east-west direction in the Earth's magnetic field. The number of neutral points,where the magnetic field is zero,is

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$A$ current of $\left(\frac{2}{\sqrt{3}}\right) \text{ A}$ produces a deflection of $60^{\circ}$ in a tangent galvanometer. The reduction factor is:

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The time period of oscillation of a magnet is $2 \ s$. When it is remagnetised so that its pole strength is $4$ times,its period will be ..... $s$.

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If the number of turns and the radius of the cross-section of the coil of a tangent galvanometer are both doubled,then the reduction factor $K$ will become:

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The unit of the reduction factor of a tangent galvanometer is

Similar Questions

Two tangent galvanometer coils of the same radius are connected in series. The current flowing produces deflections of $60^o$ and $45^o$. The ratio of the number of turns in the coils is:

$A$ bar magnet is placed on a table along the east-west direction in the Earth's magnetic field. The number of neutral points,where the magnetic field is zero,is

$A$ current of $\left(\frac{2}{\sqrt{3}}\right) \text{ A}$ produces a deflection of $60^{\circ}$ in a tangent galvanometer. The reduction factor is:

The time period of oscillation of a magnet is $2 \ s$. When it is remagnetised so that its pole strength is $4$ times,its period will be ..... $s$.

If the number of turns and the radius of the cross-section of the coil of a tangent galvanometer are both doubled,then the reduction factor $K$ will become:

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