The resistance of a circular coil is 50 turns & 10cm diameter is 5 Omega What must be the potential difference across the ends of the coil so as to multify the horizontal component of the earth's magnetic field [(B_(H)=pixx10^(-5)T] at the centre of the coil? How should the coil be placed to achieve this result?

The resistance of a circular coil is 50 turns & 10cm diameter is 5 Ome

1.	The resistance of a circular coil is 50 turns & 10cm diameter is 5 Omega What must be the potential difference across the ends of the coil so as to multify the horizontal component of the earth's magnetic field [(B_(H)=pixx10^(-5)T] at the centre of the coil? How should the coil be placed to achieve this result?
Answer» `0.5` V with PLANE of COIL NORMAL to the magnetic meridian `0.5` V with plane of coil in the magnetic meridian `0.25` V with plane of coil normal to the magnetic meridian `0.25` V with plane of the coil in the magnetic meridian Solution :N//A

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