Consider the circular loop carrying current $i$ as shown in the figure. The magnetic field at the central point $O$ is

  • A
    $\frac{2 \mu_0 i }{3 \pi R }$ acting downward
  • B
    $\frac{5 \mu_0 i }{12 R }$ acting downward
  • C
    $\frac{6 \mu_0 i }{11 R }$ acting downward
  • D
    $\frac{3 \mu_0 i }{7 R }$ acting upward

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$A$ horizontal overhead powerline is at a height of $4\,m$ from the ground and carries a current of $100\,A$ from east to west. The magnetic field directly below it on the ground is $\left( {{\mu _0} = 4\pi \times {{10}^{ - 7}}} \right)TmA^{-1}$.

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$A$ current $I=5 \text{ A}$ flows along a thin wire shaped as shown in the figure. The radius of the curved part of the wire is $R=100 \text{ mm}$, and the angle $2\phi=90^{\circ}$. The magnitude of the magnetic field at point $O$ is approximately:
$\left[\text{Use, } \frac{\mu_0}{4\pi}=10^{-7} \text{ T m A}^{-1}\right]$ (in $\mu\text{T}$)

For the adjoining figure,the magnetic field at point $P$ will be:

The magnetic field at the center of a current-carrying circular loop is $B_{1}$. The magnetic field at a distance of $\sqrt{3}R$ from the center on its axis is $B_{2}$,where $R$ is the radius of the loop. The value of $B_{1} / B_{2}$ is:

$A$ circular coil carrying current $I$ has a radius $R$ and magnetic field at the centre is $B$. The distance from the centre along the axis of the same coil where the magnetic field will be $\frac{B}{8}$ is

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