Question

the search coil has 200 turns and a cross sectional area of 3.5 x 10^-5 m². the search coil is placed at x = 0.070 m. show that the magnetic flux linkage through the search coil is about 5x 10^-4Wb

Answer 1

Solenoid

Magnetic field

$B=\frac{\mu_0 ni}{2x}$

$\frac{\phi}{A}=\frac{4\times3.14\times200\times i}{2\times0.070}$

$\frac{5\times10^{-4}}{3.5\times10^{-5}}=\frac{4\times3.14\times200\times i}{2\times0.070}$

$I=7.96\times10^{-4}A$

$B=\frac{\mu_0n I}{2r}$

$$B=\frac{4\times3.14\times10^{-7}\times200\times7.96\times10^{-4}}{2\times0.070}=1.42\times10^{-6}T$$

Magnetic flux

$\phi=BA$

$\phi=5\times10^{-4}Wb$