Question

Two circular coils each of 100 turns are held such that one lies in the vertical plane and the other in the horizontal plane with their centres coinciding. The radius of vertical and horizontal coils are 20 cm and 30 cm respectively. If the magnetic field at their common centre is zero, find current in each coil. Horizontal component of earth's magnetic field is 27.8 Aim and angle of dip at this place is ${30}^o$

Answer 1

To neutralize the magnetic field$,$ current in vertical ring should be such that the magnitude of magnetic field is $3.49\times {10}^{-5}T$ and current in horizontal ring should be such that the magnitude of magnetic filed is

$\frac{1}{\sqrt{3}}\times 3.49\times {10}^{-5}$

For vertical ring

$B=\frac{{\mu }_{0}NI}{2r}=\frac{{\mu }_0\times 100\times 1}{2\times 0.2}$

$3.49\times {10}^{-5}={\mu }_0\times 250I$

$I=\frac{3.49\times {10}^{-5}}{{\mu }_0\times 250}=\frac{3.49\times {10}^{-5}}{4\pi \times \times {10}^{-7}\times 250}=0.111A$

For horizontal ring

$B=\frac{{\mu }_{0}NI}{2r}=\frac{1}{\sqrt{3}}\times 3.49\times {10}^{-5}=\frac{{\mu }_0\times 100I}{2\times 0.3}$

$\Rightarrow 1=0.096A$

We get, $1=0.096A.$