Questions
A short magnet oscillates with a tirne period 0.1 s at a place where horizontal magnetic fleld is . 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
detailed solution
Correct option is C
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 sTalk to our academic expert!
Similar Questions
The time period of a freely suspended magnet is 4 seconds. If it is broken in length into two equal parts and one part is suspended in the same way, then its time period will be
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