Magnetic Effects Of Current Question 88
Question: Two concentric coplanar circular loops of radii $ r_{1} $ and $ r_{2} $ carry currents of respectively $ i_{1} $ and $ i_{2} $ in opposite directions (one clockwise and the other anticlockwise.) The magnetic induction at the centre of the loops is half that due to $ i_{1} $ alone at the centre. If $ r_{2}=2r_{1}. $ the value of $ I_{2}/I_{1} $ is [MP PET 2000]
Options:
A) 2
B) 1/2
C) 1/4
D) 1
Show Answer
Answer:
Correct Answer: D
Solution:
Magnetic field at centre due to smaller loop $ B_{1}=\frac{{\mu_{0}}}{4\pi }.\frac{2\pi i_{1}}{r_{1}} $ ….. (i) Due to Bigger loop $ B_{2}=\frac{{\mu_{0}}}{4\pi }.\frac{2\pi i_{2}}{r_{2}} $ So net magnetic field at centre $ B=B_{1}-B_{2}=\frac{{\mu_{0}}}{4\pi }\times 2\pi ( \frac{i_{1}}{r_{1}}-\frac{i_{2}}{r_{2}} ) $ According to question $ B=\frac{1}{2}\times B_{1} $
$ \Rightarrow \frac{{\mu_{0}}}{4\pi }.2\pi ( \frac{i_{1}}{r_{1}}-\frac{i_{2}}{r_{2}} )=\frac{1}{2}\times \frac{{\mu_{0}}}{4\pi }.\frac{2\pi i_{1}}{r_{1}} $ $ \frac{i_{1}}{r_{1}}-\frac{i_{2}}{r_{2}}=\frac{i_{1}}{2r_{1}}\Rightarrow \frac{i_{1}}{2r_{1}}=\frac{i_{2}}{r_{2}}\Rightarrow \frac{i_{1}}{i_{2}}=1 $ $ {r_{2}=2r_{1}} $
BETA
