Magnetic Effects Of Current Question 138
Question: A coil having N turns is wound tightly in the form of a spiral with inner and outer radii a and b respectively. When a current I passes through the coil, the magnetic field at the centre is [IIT-JEE (Screening) 2001]
Options:
A) $ \frac{{\mu_{0}}NI}{b} $
B) $ \frac{2{\mu_{0}}NI}{a} $
C) $ \frac{{\mu_{0}}NI}{2(b-a)}\ln \frac{b}{a} $
D) $ \frac{{\mu_{0}}I^{N}}{2(b-a)}\ln \frac{b}{a} $
Show Answer
Answer:
Correct Answer: C
Solution:
Number of turns per unit width $ =\frac{N}{b-a} $ Consider an elemental ring of radius x and with thickness dx Number of turns in the ring $ =dN=\frac{Ndx}{b-a} $ Magnetic field at the centre due to the ring element $ dB=\frac{{\mu_{0}}(dN)i}{2x}=\frac{{\mu_{0}}i}{2}.\frac{Ndx}{(b-a)}.\frac{1}{x} $
$ \therefore $ Field at the centre $ =\int_{{}}^{{}}{dB=\frac{{\mu_{0}}Ni}{2(b-a)}}\int_{a}^{b}{\frac{dx}{x}} $ $ =\frac{{\mu_{0}}Ni}{2(b-a)}\ln \frac{b}{a}. $
BETA
