Question

What will be the MMF required to produce a flux density of 0.88 T in an air gap of 8 mm?
1. 4790 A
2. 5600 A
3. 7040 A
4. 6788 A

Answer 1

Correct Answer - Option 2 : 5600 A

Formula Used:

1. Relation Between MMF and Magnetic field intensity (H):

MMF = HL .... (1)

Here "L" is the length of the gap

Also, MMF = Current × No. of Turns (NI)

2. Relation between Flux Density (B) and Magnetic field intensity (H):

B = μ_oμ_rH .... (2)

from equation (1) & (2),

MMF = NI = \(\frac{B}{μ_oμ_r}× L\)

Application:

Given,

B = 0.88 T

L = 8 mm

For air, μ_r = 1

From above formula,

MMF = \(\frac{0.88× 8 × 10^{-3}}{4π × 10^{-7}}=\frac{0.88× 8 × 10^{4}}{4π }=\frac{88× 8 × 10^{2}}{4π }\) [0.88 × 10⁴ = 88 × 10²]

Using value of π = 22/7

Hence,

MMF = \(\frac{88\times 8 \times 100}{4\times 22/7}=5600\ A\)