A short magnet oscillates with a tirne period 0.1 s at a place where horizontal magnetic fleld is 24 μT. A downward current of 18 A is established in a vertical wire 20 cm east of the magnet. The new time period of oscillator is
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a
0.1 s
b
0.089 s
c
0.076 s
d
0.057 s
answer is C.
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Detailed Solution
Initially, T=2πImBHFinally, T'=2πImB+BHWhere B = Magnetic field due to downward currentB=μ04π.2ia=18μT∴T'T=BHB+BH⇒T'0.1=2424+18⇒T'=0.076 s