# A magnetic needle is suspended freely so that it can rotate freely in the magnetic meridian. In order to keep it horizontal position, a weight of 0.2 g is kept on one end of the needle. - Physics

Sum

A magnetic needle is suspended freely so that it can rotate freely in the magnetic meridian. In order to keep it horizontal position, a weight of 0.2 g is kept on one end of the needle. If the pole strength of the needle is 20 Am, find the value of the vertical component of the Earth's magnetic field.

#### Solution

Data: M = 0.2 g = 2 x 10-4 kg, qm = 20 A·m, g = 9.8 m/s2

Without the added weight at one end, the needle will dip in the direction of the resultant magnetic field inclined with the horizontal. The torque due to the added weight about the vertical axis through the center balances the torque of the couple due to the vertical component of the Earth's magnetic field.

∴ ("Mg")("L"/2) = ("q"_"m" "B"_"v")"L"

The vertical component of the Earth's magnetic field,

"B"_"v" = "Mg"/(2"q"_"m") = ((2 xx 10^-4)(9.8))/(2(20)) = 4.9 xx 10^-5 "T"

Concept: Torque Acting on a Magnetic Dipole in a Uniform Magnetic Field
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#### APPEARS IN

Balbharati Physics 12th Standard HSC Maharashtra State Board
Chapter 11 Magnetic Materials
Exercises | Q 10 | Page 264