Q. A circular coil of 5 turns and of 10 cm mean diameter is connected to a voltage source. If the resistance of the coil is $10\ \Omega,$ the voltage of the source so is to nullify the horizontal component of earth’s magnetic field of 30 A turn $m^{- 1}$ at centre of the coil should be
A) 6 V, plane of the coil normal to magnetic meridian
B) 2 V, plane of the coil normal to magnetic meridian
C) 6 V, plane of the coil along the magnetic meridian
D) 2 V, plane of the coil along the magnetic meridian
Explanation
(a) Magnetic field of 1 A turn $m^{- 1}$
$\ \ \ \ \ \ \ \ \ \ = 4\pi \times 10^{- 7}\ T$
Field at centre $B = \frac{\mu_{0}NI}{2r} = \frac{\mu_{0}N}{2r}\ \times \frac{V}{R}$
or $V = \frac{2rRB}{\mu_{0}N}$
$\therefore\ \ \ V = \frac{2 \times \left( 5 \times 10^{- 2} \right) \times 10 \times \left( 30 \times 4\pi \times 10^{- 7} \right)}{\left( 4\pi \times 10^{- 7} \right) \times 5}$
or $V = 6\ volt$
To nullify the horizontal component of magnetic field of earth, plane of the coil should be normal to magnetic meridian.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.