In a square loop $PQRS$ made with a wire of cross-section current $i$ enters from point $P$ and leaves from point $S.$ The magnitude of magnetic field induction at the centre $O$ of the square is [image:4_image2.emf]

Question

Dr. Abubacker · Accepted Answer

(d) The resistance of arm $PQRS$ is 3 times the resistance of arm $PS$. If resistance of arm $PS = r$, then resistance of arm $PQRS = 3\ r.$
[image:4_image91.emf]
Potential difference across $P\ \ and\ S = i_{1}\ r = i_{2} 	imes 3r$
Magnetic field induction at $O$ due to current through arm $PS\ is,\ \ B_{1} = \frac{\mu_{0}}{4\pi}\frac{i_{1}}{a}\lbrack\sin{45{^\circ} + \sin{45{^\circ}brack\ }\ }$acting perpendicular to the loop upwards.
May field due to $PQ\ \ and\ RS$ are equal and opposite. Therefore natural each other magnetic field due to $QR$.
$B_{2} = \frac{\mu_{0}}{4\pi}.\frac{I_{1}}{a}\ \lbrack\sin{45{^\circ} + \sin{45{^\circ}brack\ }}$
Perpendicular to the loop in downwards.
$	herefore\ \ $Resultant magnetic field at centre $\overrightarrow{\mathbf{B}} = {\overrightarrow{\mathbf{B}}}_{1} + {\overrightarrow{\mathbf{B}}}_{2} = 0$

Q. In a square loop $PQRS$ made with a wire of cross-section current $i$ enters from point $P$ and leaves from point $S.$ The magnitude of magnetic field induction at the centre $O$ of the square is

Explanation

Reviewed by Dr. Abubacker

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