All straight wires are very long- Both AB and CD are arcs of the same circle- both subtending right angles the centre O- Then the magnetic field O is-frac - mu - - 0 - mathbf - i - - - 4 pi R -frac - mu - - 0 - mathbf - i - - - 4 pi R - sqrt - 2 -frac - mu - - 0 - mathbf - i - - - 2 pi R -frac - mu - - 0 - i - - 2 pi R - - pi - 1 -

Question

Toppr Admin · Accepted Answer

B-x3BC-0i4-x3C0-R-x3BC-0i4-x3C0-R-x3BC-0i-x3C0-2-4-x3C0-R-x2212-x3BC-0i-x3C0-2-4-x3C0-R-x3BC-0i2-x3C0-RHence-option -C- is correct answer-

All straight wires are very long. Both $A B$ and $C D$ are arcs of the same circle, both subtending right angles at the centre $O$ . Then the magnetic field at $O$ is-

$\frac{μ_{0} i}{4 π R}$
$\frac{μ_{0} i}{4 π R} \sqrt{2}$
$\frac{μ_{0} i}{2 π R}$
$\frac{μ_{0} i}{2 π R} (π + 1)$

$\begin{matrix} B = \frac{μ_{0} i}{4 π R} + \frac{μ_{0} i}{4 π R} + \frac{μ_{0} i (\frac{π}{2})}{4 π R} - \frac{μ_{0} i (\frac{π}{2})}{4 π R} = \frac{μ_{0} i}{2 π R} \end{matrix}$
Hence,
option $(C)$ is correct answer.

All straight wires are very long. Both AB and CD are arcs of the same circle, both subtending right angles at the centre O. Then the magnetic field at O is-μ0i4πRμ0i4πR√2μ0i2πRμ0i2πR(π+1)

B=μ0i4πR+μ0i4πR+μ0i(π2)4πR−μ0i(π2)4πR=μ0i2πRHence,option (C) is correct answer.

All straight wires are very long. Both $A B$ and $C D$ are arcs of the same circle, both subtending right angles at the centre $O$ . Then the magnetic field at $O$ is-

$\frac{μ_{0} i}{4 π R}$
$\frac{μ_{0} i}{4 π R} \sqrt{2}$
$\frac{μ_{0} i}{2 π R}$
$\frac{μ_{0} i}{2 π R} (π + 1)$

$\begin{matrix} B = \frac{μ_{0} i}{4 π R} + \frac{μ_{0} i}{4 π R} + \frac{μ_{0} i (\frac{π}{2})}{4 π R} - \frac{μ_{0} i (\frac{π}{2})}{4 π R} = \frac{μ_{0} i}{2 π R} \end{matrix}$
Hence,
option $(C)$ is correct answer.