Magnetism Question 47
Question: A short magnet oscillates with a time period 0.1 s at a place where horizontal magnetic field is $ 24\mu T. $ A downward current of 18 A is established in a vertical wire 20 cm east of the magnet. The new time period of oscillator
Options:
A) 0.1 s
B) 0.089 s
C) 0.076 s
D) 0.057 s
Show Answer
Answer:
Correct Answer: C
Solution:
Initially $ T=2\pi \sqrt{\frac{I}{mB _{H}}} $ , Finally $ {T}’=2\pi \sqrt{\frac{I}{m(B+B _{H})}} $ Where B = Magnetic field due to down ward conductor $ =\frac{{\mu _{0}}}{4\pi }.\frac{2i}{a}=18\mu T $ \ $ \frac{{{T}’}}{T}=\sqrt{\frac{B _{H}}{B+B _{H}}} $
Therefore $ \frac{{{T}’}}{0.1}=\frac{24}{18+24} $
Therefore $ {T}’=0.076,s. $
BETA
