Question

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 as to nullify the horizontal component of earths magnetic field of $ 30\, A $ 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.