Question

The value of the magnetic field induction at the centre of the coil, if l = 2A, h= 10 cm. in the following figure is

• A. $1.14\times10^{-5}\,T$
• B. $\frac{4\pi}{\mu_0}$ $1.14\times10^{-5}\,T$
• C. $\frac{1.14}{4\pi}$ $\times10^{-5}\,T$
• D. $\frac{\mu_0}{4\pi}$ $2.14\times10^{-5}\,T$

Answer 1

Answer: (A)

$1.14\times10^{-5}\,T$

Answer 2

For the upper straight wire, $B_1$ = 0, $\because$ $\theta$ =0 For the lower straight wire. $B_2 = \frac{\mu_0}{4 \,\pi} \left[ \frac{I}{h} \right]$ $\because$ $\phi_1$ = 90, $\phi_2$ = 0 For circular portion, $B_3 = \frac{\mu_0 }{4 \, \pi} \left[ \frac{3 \, \pi \, I}{2 \, h} \right]$ $\because$ $\beta = \frac{3 \,\pi}{2}$ and $r = h$ Hence total field, $B = \frac{\mu_0}{4 \, \pi} \left[ \frac{I}{h} \left( \frac{3\, \pi}{2} + 1 \right) \right] = 1.14 \times 10^{-5}$ T Perpendicular to the plane of the figure and directed outward