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Question
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. (g = 9.8 ms-2)
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Solution
Given: Here mass m = 0 .2 g = 0 .2 x 10-3 kg, g = 9.8 ms-2 Pole strength = 20 Am
To find: B =Earth's magnetic field=?
Now, Let mg = pole strength x B
∴ `0.2 xx 10^-3 xx 9.8 = 20 xx B`
∴ `(0.2 xx 10^-3 xx 9.8) /20`
B = 9 .8 x 10-5 T
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