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A 2 turn coil of radius 10cm is placed w...

A 2 turn coil of radius `10cm` is placed with its plane in magnetic meridian. A small magnetic needle is suspended at the centre of the coil by a torsion free silk thread. On passing a current through the coil, the needle is deflected through `45^@`. Calculate the strength of the current if horizontal component of earth's field is `1*6xx10^-5T`

Text Solution

Verified by Experts

The correct Answer is:
`1.27`
Here , n = 2 , r= 10 cm = 0.1 m , `theta = 45^(@) ` I = ?
`H = 1.6 xx 10^(-5) T`
As `I = (2rH)/(mu_(0)n) tan theta `
`:. I = (2xx0.1 xx 1.6 xx 10^(-5) tan45^(@))/(4pi xx 10^(-7) xx2)=1.27 A`
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