(B) The principle of a tangent galvanometer is given by the formula: $B = B_H \tan \theta$,where $B$ is the magnetic field produced by the coil,$B_H$

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(B) The principle of a tangent galvanometer is given by the formula: $B = B_H \tan \theta$,where $B$ is the magnetic field produced by the coil,$B_H$ is the horizontal component of the Earth's magnetic field,and $\theta$ is the deflection angle.Given:$B_H = 0.34 \times 10^{-4} \, T$$\theta = 30^{\circ}$Substituting the values:$B = (0.34 \times 10^{-4}) \times \tan(30^{\circ})$Since $\tan(30^{\circ}) = \frac{1}{\sqrt{3}} \approx 0.57735$$B = 0.34 \times 10^{-4} \times 0.57735$$B \approx 0.1963 \times 10^{-4} \, T$$B \approx 1.963 \times 10^{-5} \, T$Thus,the magnetic field of the coil is $1.96 \times 10^{-5} \, T$.

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

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$A$ tangent galvanometer shows a deflection of $30^{\circ}$ under the influence of the horizontal component of the Earth's magnetic field,which is $0.34 \times 10^{-4} \, T$. What is the magnetic field of the coil?

A
$1.96 \times 10^{-4} \, T$
B
$1.96 \times 10^{-5} \, T$
C
$1.96 \times 10^{4} \, T$
D
$1.96 \times 10^{5} \, T$

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

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MediumAP EAMCET 2022

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$A$ tangent galvanometer has a coil of $100$ turns and a radius of $20\, cm$. The horizontal component of the earth's magnetic field is $B_H = 3 \times 10^{-5}\, T$. Find the current which gives a deflection of $45^{\circ}$. (in $, A$)

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DifficultTS EAMCET 2008

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The time period of a freely suspended magnet is $2 \ s$. If it is broken in length into two equal parts and one part is suspended in the same way,then its time period will be.....$s$.

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$A$ tangent galvanometer shows a deflection of $30^{\circ}$ under the influence of the horizontal component of the Earth's magnetic field,which is $0.34 \times 10^{-4} \, T$. What is the magnetic field of the coil?

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