Question

An element of $0.05\, \hat{i} \,m$ is placed at the origin as shown in figure which carries a large current of $10 \,A$. The magnetic field at a distance of $1 \,m$ in perpendicular direction is

• A. $4.5 \times 10^{-8}\, T$
• B. $5.5 \times 10^{-8}\, T$
• C. $5.0 \times 10^{-8}\,T$
• D. $7.5 \times 10^{-8}\, T$

Answer 1

Answer: (C)

$dB=\frac{\mu_{0}}{4\pi} \frac{Idl \,sin\,\theta}{r^{2}}$
Here, $dl=\Delta x=0.05\,m, I=10\,A, r=1\,m$
$sin\theta=sin\, 90^{\circ}=1$,
$\therefore dB=10^{-7}\times\frac{10\times0.05\times1}{\left(1\right)^{2}}$
$=0.50\times10^{-7}$
$=5.0\times10^{-8}\,T$