Question

A compass needle free to turn in a horizontal plane is placed at the centre of a circular coil of $30$ turns and radius $12 \,cm$. The coil is in a vertical plane making an angle of $45^{\circ}$ with the magnetic meridian when the current in the coil is $0.35 \,A$, the needle points west to east. Determine the horizontal component of earth's magnetic field at the location.

• A. $3.9 \times 10^{-7}$ tesla
• B. $3.9 \times 10^{-5}$ tesla
• C. $8.0 \times 10^{-5}$ tesla
• D. $7.0 \times 10^{-7}$ tesla

Answer 1

Answer: (B)

$3.9 \times 10^{-5}$ tesla

Answer 2

Here, $n = 30$, $r = 12 \,cm$ $= 12 \times 10^{-2}\, m$. $i = 0.35 \,A$, $H=?$ As it is clear from figure shown the needle can point west to east only when $H = B \,sin\, 45^{\circ}$ where, $B =$ magnetic field strength due to current in coil $= \frac{\mu_{0}}{4\pi} \frac{2\pi ni}{r}$ $\therefore\quad H = \frac{\mu _{0}}{4\pi } \frac{2\pi ni}{r} \,sin\, 45^{\circ}$ $= 10^{-7} \times \frac{2\pi\times 30\times 0.35}{12 \times 10^{-2}} \cdot \frac{1}{\sqrt{2}}$ $= 2\times \frac{22}{7}\times \frac{30\times 35}{12\times \sqrt{2}} \times 10^{-7}$ $= 3.9 \times 10^{-5}$ tesla