Question

A circular coil of $5$ turns and of $10cm$ mean diameter is connected to a voltage source. If the resistance of the coil is $10\Omega$ the voltage of the source so as to nullify the horizontal component of earths magnetic field of $30A$ turn $m^{-1}$ at the 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
• E. 4V, plane of the coil normal to magnetic meridian

Answer 1

Answer: (A)

Magnetic field of 1 A turn/m $=4\pi \times {10}^{-7}T$
Field at centre $B=\frac{{\mu }_{0}NI}{2r}=\frac{{\mu }_{0}N}{2r}\times \frac{V}{R}$
Or $V=\frac{2rRB}{{\mu }_{0}N}$
$\therefore$ $B=\frac{2\times (5\times {10}^{-2})\times 10\times (30\times 4\pi \times {10}^{-7})}{(4\pi \times {10}^{-7})\times 5}$
or $V=6$
volt To nullify the horizontal component of magnetic field of earth, plane of the coil should be normal to magnetic meridian.