Electromagnetic Induction And Alternating Current Ques 18
- A circular loop of radius $0.3 cm$ lies parallel to a much bigger circular loop of radius $20 cm$. The centre of the smaller loop is on the axis of the bigger loop. The distance between their centres is $15 cm$. If a current of $2.0 A$ flows through the bigger loop, then the flux linked with smaller loop is
(2013 Main)
(a) $9.1 \times 10^{-11} Wb$
(b) $6 \times 10^{-11} Wb$
(c) $3.3 \times 10^{-11} Wb$
(d) $6.6 \times 10^{-9} Wb$
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Answer:
Correct Answer: 18.(a)
Solution:
Formula:
- Magnetic field at the centre of smaller loop
$$ B=\frac{\mu _0 i R _2^{2}}{2\left(R _2^{2}+x^{2}\right)^{3 / 2}} $$
Area of smaller loop $S=\pi R _1^{2}$
$\therefore$ Flux through smaller loop $\varphi=B S$
Substituting the values, we get, $\varphi \approx 9.1 \times 10^{-11} Wb$
BETA
