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A compass needle free to turn in a horizontal plane is placed at the centre of circular coil of 30 turns and radius $12 \mathrm{~cm}$. The coil is in a vertical plane making an angle of $45^{\circ}$ with the magnetic meridian. When the current in the coil is $0.35 \mathrm{~A}$, the needle points west to east.
(a) Determine the horizontal component of the earth's magnetic field at the location.
(b) The current in the coil is reversed, and the coil is rotated about its vertical axis by an angle of $90^{\circ}$ in the anticlockwise sense looking from above. Predict the direction of the needle. Take the magnetic declination at the places to be zero.
(a) Determine the horizontal component of the earth's magnetic field at the location.
(b) The current in the coil is reversed, and the coil is rotated about its vertical axis by an angle of $90^{\circ}$ in the anticlockwise sense looking from above. Predict the direction of the needle. Take the magnetic declination at the places to be zero.
Solution:
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Verified Answer
(a) Given, for the coil,
$$
\mathrm{N}=30, \mathrm{r}-12 \mathrm{~cm}=0.12 \mathrm{~m}, 1=0.35 \mathrm{~A}
$$
At center, $B=\frac{\mu_0 I N}{2 r}$
Now $\mathrm{B}$ is inclined to $\mathrm{B}_{\mathrm{H}}$ by $135^{\circ}$ (being $90^{\circ}$ to plane of coil) and resultant is at $90^{\circ}$ with $\mathrm{B}_{\mathrm{H}}$ (needle pointing from west to east)
From parallelogram law formula,
$$
\tan 90^{\circ}=\frac{B \sin 135^{\circ}}{B_H+B \cos 135^{\circ}}
$$
or $\infty=\frac{B \sin 45^{\circ}}{B_H-\cos 45^{\circ}}$ i. e., $B_{\mathrm{H}}-\mathrm{B} \cos 45^{\circ}=0$
$$
\begin{aligned}
&\text { or, } \mathrm{B}_{\mathrm{H}}=\mathrm{B} \cos 45^{\circ}=\frac{\mu_0 I N}{2 r} \cos 45^{\circ} \\
&=\frac{4 \pi \times 10^{-7} \times 0.35 \times 30 \times 0.707}{2 \times 0.12} \\
&=\frac{4 \times 22 \times 35 \times 3 \times 707}{7 \times 2 \times 12} \times 10^{-9} \\
&=11 \times 5 \times 707 \times 10^{-9}=38885 \times 10^{-9} \mathrm{~T}=0.38885 \mathrm{G} .
\end{aligned}
$$
(b) When current is reversed and coil rotated by $90^{\circ}$, resultant field is reversed. It points east to west, The needle will also stay east to west
$$
\mathrm{N}=30, \mathrm{r}-12 \mathrm{~cm}=0.12 \mathrm{~m}, 1=0.35 \mathrm{~A}
$$
At center, $B=\frac{\mu_0 I N}{2 r}$
Now $\mathrm{B}$ is inclined to $\mathrm{B}_{\mathrm{H}}$ by $135^{\circ}$ (being $90^{\circ}$ to plane of coil) and resultant is at $90^{\circ}$ with $\mathrm{B}_{\mathrm{H}}$ (needle pointing from west to east)
From parallelogram law formula,
$$
\tan 90^{\circ}=\frac{B \sin 135^{\circ}}{B_H+B \cos 135^{\circ}}
$$
or $\infty=\frac{B \sin 45^{\circ}}{B_H-\cos 45^{\circ}}$ i. e., $B_{\mathrm{H}}-\mathrm{B} \cos 45^{\circ}=0$
$$
\begin{aligned}
&\text { or, } \mathrm{B}_{\mathrm{H}}=\mathrm{B} \cos 45^{\circ}=\frac{\mu_0 I N}{2 r} \cos 45^{\circ} \\
&=\frac{4 \pi \times 10^{-7} \times 0.35 \times 30 \times 0.707}{2 \times 0.12} \\
&=\frac{4 \times 22 \times 35 \times 3 \times 707}{7 \times 2 \times 12} \times 10^{-9} \\
&=11 \times 5 \times 707 \times 10^{-9}=38885 \times 10^{-9} \mathrm{~T}=0.38885 \mathrm{G} .
\end{aligned}
$$
(b) When current is reversed and coil rotated by $90^{\circ}$, resultant field is reversed. It points east to west, The needle will also stay east to west
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