Magnetic Effects Of Current Question 132
Question: The field normal to the plane of a wire of n turns and radius r which carries a current i is measured on the axis of the coil at a small distance h from the centre of the coil. This is smaller than the field at the centre by the fraction
Options:
A) $ \frac{3}{2}\frac{h^{2}}{r^{2}} $
B) $ \frac{2}{3}\frac{h^{2}}{r^{2}} $
C) $ \frac{3}{2}\frac{r^{2}}{h^{2}} $
D) $ \frac{2}{3}\frac{r^{2}}{h^{2}} $
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Answer:
Correct Answer: A
Solution:
Field at the centre $ B_{1}=\frac{{\mu_{0}}}{4\pi }\times \frac{2\pi in}{r}=\frac{{\mu_{0}}}{2}.\frac{ni}{r} $ Field at a distance h from the centre $ B_{2}=\frac{{\mu_{0}}}{4\pi }.\frac{2\pi nir^{2}}{{{(r^{2}+h^{2})}^{3/2}}}=\frac{{\mu_{0}}}{2}.\frac{nir^{2}}{r^{3}{{( 1+\frac{h^{2}}{r^{2}} )}^{3/2}}} $ $ =B_{1}{{( 1+\frac{h^{2}}{r^{2}} )}^{-3/2}}=B_{1}( 1-\frac{3}{2}.\frac{h^{2}}{r^{2}} ) $ (By binomial theorem) Hence B2 is less than B1 by a fraction $ =\frac{3}{2}\frac{h^{2}}{r^{2}} $
BETA
