Question

A tightly wound $90$ turn coil of radius $15$ cm has a magnetic field of $4\times {10}^{-4}T$ at its center. The current flowing through it is :

• A. $1.06A$
• B. $2.44A$
• C. $3.44A$
• D. $4.44A$

Answer 1

The correct option is A $1.06A$

Here, $N=90$
$R=15cm=15\times {10}^{-2}m,B=4\times {10}^{-4}T$
$\because B=\frac{{\mu }_{o}NI}{2R}$
$\therefore I=\frac{2RB}{N{\mu }_o}=\frac{2\times 15\times {10}^{-2}\times 4\times {10}^{-4}}{4\pi \times {10}^{-7}\times 90}=1.06A$.