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Magnetism and Matter question

2010 · Q146
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Magnetism and Matter question

2010 · Q146

NEETPhysicsMagnetism and MatterMCQ+4 / −1
A vibration magnetometer placed in magnetic meridian has a small bar magnet. The magnet executes oscillations with a time period of 2 sec in earth's horizontal magnetic field of 24 microtesla. When a horizontal field of 18 microtesla is produced opposite to the earth's field by placing a current carrying wire, the new time period of magnet will be
  1. A
    1 s
  2. B
    2 s
  3. C
    3 s
  4. D
    4 s
View written solutionFree

Correct answer: D

The time period T of oscillation of a magnet is given by

T = 2πIMB2\pi \sqrt {{I \over {MB}}} 2πMBI​​

As I, B remains the same here

So T ∝\propto∝ 1B{1 \over {\sqrt B }}B​1​

or T2T1=B1B2{{{T_2}} \over {{T_1}}} = \sqrt {{{{B_1}} \over {{B_2}}}} T1​T2​​=B2​B1​​​

⇒\Rightarrow⇒ T2 = T1B1B2{T_1}\sqrt {{{{B_1}} \over {{B_2}}}} T1​B2​B1​​​

= 224×10−66×10−62\sqrt {{{24 \times {{10}^{ - 6}}} \over {6 \times {{10}^{ - 6}}}}} 26×10−624×10−6​​ = 4 s

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