Sum

A tangent galvanometer shows a deflection of 45° when 10 mA of current is passed through it. If the horizontal component of the earth's magnetic field is `B_H = 3.6 xx 10^-5 "T"` and radius of the coil is 10 cm, find the number of turns in the coil.

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#### Solution

Given :

Horizontal component of Earth's magnetic field,`B_H = 3.6 xx 10^-5 "T"`

Deflection shown by the tangent galvanometer, `θ = 45^circ`

Current through the galvanometer, `l = 10 "mA" = 10^-2 "A"`

Radius of the coil, r = `10 "cm" = 0.1 "m"`

Number of turns in the coil, n = ?

We know ,

`B_H tan θ = (u_0 In)/(2r)`

⇒ `n = (B_H tan \ θ xx 2r)/(u_0I)`

⇒ `n = (3.6 xx 10^-5 xx 2 xx 1 xx 10^-1)/(4pi xx 10^-7 xx 10^-2)`

⇒ `n = 0.57332 xx 10^3 = 573`

Concept: Magnetic Properties of Materials

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