A magnetic needle is arranged at the centre of a current carrying coil having 50 turns with radius of coil ….

Q: A magnetic needle is arranged at the centre of a current carrying coil having 50 turns with radius of coil 20 cm arranged along magnetic meridian. When a current of 0.5 mA is allowed to pass through the coil the deflection is observed to be 30°. Find the horizontal component of earth’s magnetic field.

Sol: $\large B = B_H tan\theta $

$\large \frac{\mu_0 n i}{2 r} = B_H tan\theta $

$\large B_H = \frac{\mu_0 n i}{2 r tan \theta} $

$ \large B_H = \frac{4 \pi \times 10^{-7} \times 50 \times 5 \times 10^{-4}}{2 \times 0.2 \times tan30^o} $

$\large = 5 \sqrt{3} \pi \times 10^{-8} T$

= 26.35  × 10-8 T

= 2.365  × 10-7 T