The magnetic needle of a tangent galvanometer | vedclass

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Question

(B) In a tangent galvanometer,the magnetic field $B_{m}$ produced by the magnet (or current in the coil) is related to the horizontal component of the

Vedclass · Accepted Answer

$1.96 	imes 10^{-5} 	ext{ T}$

Vedclass · Answer

(B) In a tangent galvanometer,the magnetic field $B_{m}$ produced by the magnet (or current in the coil) is related to the horizontal component of the Earth's magnetic field $B_{H}$ by the formula:$B_{m} = B_{H} 	an(	heta)$Given:$B_{H} = 0.34 	imes 10^{-4} 	ext{ T}$$	heta = 30^{\circ}$Substituting the values:$B_{m} = (0.34 	imes 10^{-4} 	ext{ T}) 	imes 	an(30^{\circ})$Since $	an(30^{\circ}) = \frac{1}{\sqrt{3}} \approx 0.57735$$B_{m} = 0.34 	imes 10^{-4} 	imes 0.57735 	ext{ T}$$B_{m} \approx 0.1963 	imes 10^{-4} 	ext{ T} = 1.963 	imes 10^{-5} 	ext{ T}$Thus,the correct option is $B$.

The magnetic needle of a tangent galvanometer is deflected at an angle $30^{\circ}$ due to a magnet. The horizontal component of the Earth's magnetic field is $0.34 \times 10^{-4} \text{ T}$,which is along the plane of the coil. The magnetic field due to the magnet is:

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