Question

Two mutually perpendicular concentric current carrying coils have the same radius $2\pi cm$ . If currents $3A$ and $4A$ flow in each coil respectively, then calculate the magnitude of magnetic induction (in $Wb/m^2$ ) at the common centre of the two coils. $\left({\left(\mu \right)}_0=4\pi \times {\left(10\right)}^{-7}WbA{m}^{-1}\right)$

• A. $7\times 10^{-5}$
• B. $5\times 10^{-5}$
• C. $10^{-5}$
• D. $12\times 10^{-5}$

Answer 1

Answer: (B)

The magnetic field at the centre of a circular loop is given by;
$B=\frac{{\mu }_{0}i}{2r}$
Thus,
$B_1=\frac{{\left(\mu \right)}_0{i}_1}{2r}=\frac{{\left(\mu \right)}_o{i}_1}{2\left(2\pi \times {\left(10\right)}^{-2}\right)}=\frac{{\left(\mu \right)}_o\times 3\times {\left(10\right)}^2}{4\pi }$
$B_2=\frac{{\left(\mu \right)}_0{i}_2}{2r}=\frac{{\left(\mu \right)}_o{i}_2}{2\left(2\pi \times {\left(10\right)}^{-2}\right)}=\frac{{\left(\mu \right)}_o\times 4\times {\left(10\right)}^2}{4\pi }$
Since $B_1$ and $B_2$ are perpendicular to each other, resultant magnetic field is;
$B_{net}=\sqrt{B_1^2+B_2^2}=\frac{{\mu }_o}{4\pi }5\times 10^2$
$B_{net}=10^{-7}\times 5\times 10^2$
$B_{net}=5\times 10^{-5}Wb/m^2$