Question

The wire shown in figure carries a current of $40A$. If $r=3.14cm$ the magnetic field at point p will be :

• A. $1.6\times {10}^{-3}T$
• B. $3.2\times {10}^{-3}T$
• C. $6\times {10}^{-4}T$
• D. $4.8\times {10}^{-3}T$

Answer 1

Answer: (C)

$6\times {10}^{-4}T$

Magnetic field due to circular arc is given by,
$\left|B\right|=\frac{{\mu }_{0}I}{4\pi r}\theta$ =$\frac{{\mu }_{0}I\times 3\pi }{4\pi r\times 2}$=$\frac{3{\mu }_{0}I}{8r}$
=$\frac{3\times 4\pi \times {10}^{-7}\times 40}{8\times 3.14\times {10}^{-2}}$
=$6\times {10}^{-4}T$