Question

A tangent galvanometer has 80 turns of wire. The internal and external diameters of the coil are 19 cm and 21 cm respectively. The reduction factor of the galvanometer at a place where H = 0.32 oersted will be (1 oersted = 80 A/m) is 0.064.How?

Answer 1

formula of reduction factor is given by
$\bold{K=\frac{2rB}{\mu_0 N}}$
Where r is the average value of inner and outer radii of coil
B is the magnetic field , N is the number of turns of wire in coil

Here,
H = 0.32 oersted = 0.32 × 80 A/m but B = μ₀H =μ₀ 0.32 × 80 Tesla
N = 80
r = (r₁ + r₂)/2 ⇒ r₁ = 19/2 cm and r₂ = 21cm
r = (19/2 + 21/2)/2 = 10cm = 10 × 10⁻² m = 0.1 m

Now, reduction factor ( R ) = 2 × 0.1 × μ₀ 0.32 × 80/μ₀80 = 2 × 0.1 × 0.32 = 0.064

Hence, reduction factor is 0.064

Answer 2

we have the formula,
K = (2rB/u•n) - - - -(1)
so,
H = 0.3 x 80
B = u•H => u• x (0.3 x 80)
n = 80
r = (r¹ + r²)/2 = (19+21)/2 = 0.1 m
Putting all values in (1), we get,
K = 0.064