The magnetic field $d\overrightarrow B $ due to a small current element $d\overrightarrow {l\,} $ at a distance $\overrightarrow {r\,} $ and element carrying current $i$ is
$d\overrightarrow B = \frac{{{\mu _0}}}{{4\pi }}i\,\left( {\frac{{d\overrightarrow {l\,} \times \overrightarrow {r\,} }}{r}} \right)$
$d\overrightarrow B = \frac{{{\mu _0}}}{{4\pi }}{i^2}\,\left( {\frac{{d\overrightarrow {l\,} \times \overrightarrow {r\,} }}{r}} \right)$
$d\overrightarrow B = \frac{{{\mu _0}}}{{4\pi }}{i^2}\,\left( {\frac{{d\overrightarrow {l\,} \times \overrightarrow {r\,} }}{{{r^2}}}} \right)$
$d\overrightarrow B = \frac{{{\mu _0}}}{{4\pi }}i\,\left( {\frac{{d\overrightarrow {l\,} \times \overrightarrow {r\,} }}{{{r^3}}}} \right)$
A current $I$ flows in an infinitely long wire with cross section in the form of a semi-circular ring of radius $R$. The magnitude of the magnetic induction along its axis is:
A thin ring of $10\, cm$ radius carries a uniformly distributed charge. The ring rotates at a constant angular speed of $40\,\pi \,rad\,{s^{ - 1}}$ about its axis, perpendicular to its plane. If the magnetic field at its centre is $3.8 \times {10^{ - 9}}\,T$, then the charge carried by the ring is close to $\left( {{\mu _0} = 4\pi \times {{10}^{ - 7}}\,N/{A^2}} \right)$
One Tesla is equal to
In the figure shown there are two semicircles of radii ${r_1}$ and ${r_2}$ in which a current $i$ is flowing. The magnetic induction at the centre $O$ will be
Find magnetic field at point $P$ in given diagram.