The resistance of a circular coil of 50 turns and 10 cm diameter is `5Omega`. What must be the potential difference across the ends of the coil so as to nullify the earth's magnetic field (H=0.314 gauss) at the centre of the coil? How should the coil be placed to achieve this result?

Options

A. 0.5 V with plane of coil normal to the magnetic meridian
B. 0.5 V with plane of ocil in the magnetic meridian
C. 0.25 V with plane of coil normal to the magnetic meridian
D. 0.25 V with plane of coil in the magnetic meridian

Answer

C

Explanation

Given, `n=50,r=(10)/(2)=5cm=5xx10^(-2)m`
`R=5Omega and H=0.314` gauss
For ciruclar coil `H=(mu_(0))/(4pi)(2pin i)/(r)`
or `i=(4pi)/(mu_(0))(rH)/(2pin)`
`=(5xx10^(-2)xx0.314xx10^(-4))/(10^(-6)xx2xx3.14xx50)`
`=5xx10^(-2)A`
`V=iR`
`=5xx10^(-2)xx5=0.25V`
Magnetic field produced due to the coil is normal to the plane of coil so that plane of the coil should be normal to the magnetic meridian.
Hence choice (c) is correct.