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
Correct Answer: A
Solution :
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.
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