Question

Two coils each of 100 turns are held such that one lies in vertical plane and the other in the horizontal plane with their centers coinciding. The radius of the vertical coil is $0.20m$ and that of the horizontal coil is $0.30m$. How will you neutralize the magnetic field of the earth at their common center? If the summation of current to be passed through each coil is $i_{total}=xA$. Find x*10 .

Horizontal component of earth's magnetic field$=0.35\times {10}^{-4}$ $T$ and angle of dip$={30}^o$.

Answer 1

Angle of dip $\theta ={\tan }^{-1}\left(\frac{B_V}{B_H}\right)$ where $B_V$ and $B{)}_H$ are the vertical and horizontal components of earth's field respectively.
Now, $B_V=B_H\tan {30}^o=(0.35\times {10}^{-4})\times 0.577=0.202\times {10}^{-4}T$.

It should be remembered that the direction of magnetic field will be perpendicular to the plane of either coil. Let the vertical coil be perpendicular to the magnetic meridian. The field produced by this coild can be used to neutralise the horizontal component of earth's field, if:

$B_H=\left(\frac{{\mu }_0}{4\pi }\right)\cdot \frac{2\pi n{i}_1}{r}$

$0.35\times {10}^{-4}=\left(\frac{{\mu }_0}{4\pi }\right)\cdot \frac{2\pi n{i}_1}{r}$

$\Rightarrow i_1=0.111A$

Similarly, the magnetic field produced by the horizontal coil will be vertical and will newtralise the vertical component $B_V$ if:

$B_V=\left(\frac{{\mu }_0}{4\pi }\right)\cdot \frac{2\pi n{i}_2}{r}$
$0.202\times {10}^{-4}=\left(\frac{{\mu }_0}{4\pi }\right)\cdot \frac{2\pi n{i}_2}{r}$

$\Rightarrow i_2=0.0964A$