Question

If a long straight wire carries a current of 40 A, then the magnitude of the field B at a point 15 cm away from the wire is:

• A. 5.34 x ${10}^5$ T
• B. 8.34 x ${10}^5$ T
• C. 9.6 x ${10}^5$ T
• D. 10.2 x ${10}^5$ T

Answer 1

Answer: (A)

5.34 x ${10}^5$ T

$I=40$A
$r=15cm=15\times {10}^{-2}$m

$\therefore$ $B$ = $\frac{{\mu }_{0}I}{2\pi R}=\frac{2\times {10}^{-7}\times 40}{15\times {10}^{-2}}=\frac{80}{15}\times {10}^{-5}=5.34\times {10}^{-5}T$