Question

A moving-coil instrument has 100 turns
of wire with a resistance of 10 12, an active
length in the gap of 3 cm and width of 2 cm.
Ap.d. of 45 mV produces full-scale
deflection. The control spring exerts a
torque of 490.5 x 10-7 N-m at full-scale
deflection. Calculate the flux density in the
gap.​

Answer 1

The flux density of the gap is  1.8167  × $10^{-9}$  wb/m²

Explanation:

Given as :

For A moving-coil instrument

The number of turns of wire = n = 100

The resistance of wire = 10 ohm

The length of gap = l = 3 cm = 0.03 m

The width of gap = w = 2 cm = 0.02 m

The potential difference  = v = 45 m v = 45 × $10^{-3}$  volt  = 0.045 volt

The torque of spring = 490.5 × $10^{-7}$   N-m

Let The flux density of the gap = B   wb/m²

According to question

For Torque of the spring at full scale deflection

∵ Torque = B I N A

Where B = flux density of the gap

           I = Current

           N = number of turns

          A = Area of gap

So, Current = I = $\dfrac{potential}{resistance}$

                      I = $\dfrac{0.045}{10}$   = 4.5 × $10^{-3}$  amp

Area of gap =  length of gap ×  breadth of gap

                    = 0.03 m × 0.02  m

                    = 0.6 × $10^{-3}$    m²

Now, From Torque equation

Torque = B I N A

Or,   490.5 × $10^{-7}$   =  B ×  4.5 × $10^{-3}$  × 100 × 0.6 × $10^{-3}$

Or, 490.5 × $10^{-7}$   =  B × 2.7 × $10^{-4}$

∴                      B = $\dfrac{490.5\times 10^{-7}}{2.7\times 10^{-4}}$

i.e                    B = 1.8167  × $10^{-9}$    wb/m²

So, The flux density of the gap = B =  1.8167  × $10^{-9}$   wb/m²

Hence, The flux density of the gap is  1.8167  × $10^{-9}$  wb/m²  Answer