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) \[\frac{{{\mu }_{0}}{{I}^{N}}}{2(b-a)}\ln \frac{b}{a}\]
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 \[=\int_{{}}^{{}}{dB=\frac{{{\mu }_{0}}Ni}{2(b-a)}}\int_{a}^{b}{\frac{dx}{x}}\] \[=\frac{{{\mu }_{0}}Ni}{2(b-a)}\ln \frac{b}{a}.\]You need to login to perform this action.
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