Question

The wire shown in Fig. carries a current of 40 A. If r = 3.14 cm, the magnetic field at point P will be

• A. $1.6\times {10}^{-3}T$
• B. $3.2\times {10}^{-3}T$
• C. $4.8\times {10}^{-3}T$
• D. $6.4\times {10}^{-3}T$

Answer 1

The correct option is C

$4.8\times {10}^{-3}T$

The straight portions of the wire do not contribute because the point P is along them. The field at P is due to $\frac{3}{4}th$ of the loop of radius r. Thus
$B=\frac{3}{4}\left(\frac{{\mu }_{0}I}{2r}\right)=\frac{3}{4}\times \frac{4\pi \times {10}^{-7}\times 40}{3.14\times {10}^{-2}}=4.8\times {10}^{-3}T$
Hence the correct choice is (c).