Question

The magnitude of the magnetic field at the centre of the tightly wound 150 turn coil of radius 12 cm carrying a current of 2 A is:

• A. 18 G
• B. 19.7 G
• C. 15.7 G
• D. 17.7 G

Answer 1

The correct option is C 15.7 G

Given,

$N=150$

$I=2A$

$r=12cm=0.12m$

The magnitude of the magnetic field at the center is

$B=\frac{{\mu }_{0}NI}{2r}$

$B=\frac{4\pi \times {10}^{-7}\times 150\times 2}{2\times 0.12}$

$B=\frac{2\times 600\times 3.14\times {10}^{-7}}{0.24}$

$B=15.7\times {10}^{-4}T$

$B=15.7G$