Question

A pair of stationary and infinitely long bent wires are placed in the $x-y$ plane as shown in the figure. The two wires carry currents of $4A$ and $9A$ respectively. If the segments $D$ and $A$ are along $x$-axis and the segments $C$ and $F$ are parallel to $y$-axis such that $OB=2cm$ and $OE=3cm$, then the magnitude of the magnetic induction at the origin '$O$' is: $({\mu }_0=4\pi \times {10}^{-7}H{m}^{-1})$

• A. $4\times {10}^{-5}T$
• B. $5\times {10}^{-5}T$
• C. $6\times {10}^{-5}T$
• D. $2\times {10}^{-5}T$

Answer 1

The correct option is B $5\times {10}^{-5}T$

Due to segments $D$ and $A$, respective magnetic fields are zero using Biot Savart's Law.

$B_D=0$

$B_A=0$

For a finite current carrying wire as shown in the attached figure, the magnetic field is given by $B=\frac{{\mu }_{o}I}{4\pi R}(\sin {\theta }_1+\sin {\theta }_2)$

Direction of magnetic field is given by right hand thumb rule (or alternatively from Biot-Savart Law).

Due to segment $C$, magnetic field is:

$B_C=\frac{{\mu }_o{I}_C}{4\pi R_C}(\sin {\theta }_{1,C}+\sin {\theta }_{2,C})$

$B_C=\frac{{10}^{-7}\times 4}{2\times {10}^{-2}}(\sin 0+\sin ({90}^o\left)\right)$

$B_C=2\times {10}^{-5}T$ directed outside the plane of paper

Similarly,

$B_F=3\times {10}^{-5}T$ directed outside the plane of paper

Net field at $O$ is:

$B=B_A+B_C+B_D+B_F$

$B=5\times {10}^{-5}T$ directed outside the plane of paper