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Q.
The wire shown in figure carries a current of 40A. The
radius of the circular part of the wire is 2 cm. Find the
magnetic field at the centre of the circular part of the
wire.
Solution:
Magnetic field due to current element $I \overrightarrow{d\, I}$ at O is given by $dB = \frac{\mu_0}{4 \pi } \frac{Idl}{r^2} = \frac{\mu_0}{4 \pi} \frac{Idl}{r^2} $
($\because$ sin 90$^\circ$ = 1)
$\therefore $ magnetic field at O due to the whole circular part AB is B = $\frac{\mu_0}{4 \pi} \frac{I}{r^2} \, \int\limits_0^{\frac{3}{4} \times 2 \pi r}$
= $\frac{\mu_0}{4 \pi } \frac{I}{r^2} \times \frac{3}{4} \times 2 \pi r = \frac{\mu_0}{4 \pi } \left( \frac{3I \pi}{2r} \right)$
B = $\frac{10^{-7} \times 3 \times 40 \times 3.142}{2 \times 2 \times 10^{-2}} = 9.43 \times 10^{-4}$T
