Question

A ring shaped electromagnet has an air gap of 6mm and cross sectional area of 12 cm².The mean length of the core (excluding the air gap) is 60 cm. Calculate the mmf required to produce a flux density of 0.4wb/m² in the gap? (take relative permeability of the material as 400)

anyone answer me..plsss....
​

Answer 1

application matar as400 take too proud a calculate the MS Word what is computer section get stormont ourang shaped altamont electromart Aurangzeb off meta off sr400 Rupees $400 dollar is 2 example 100000

Answer 2

Step-by-step Explanation:

Given: Length of the air gap $l_{gap}$ = 6 mm = 6 × 10⁻³ m

Length of the core ring $l_{ring}$ = 60 cm = 6 × 10⁻² m

flux density (B) = 0.4wb/m²

Relative permeability of the material $\mu_{r}$ = 400

To Find: the m.m.f. required to produce the given flux density

Solution:

• Calculating m.m.f. required to produce a flux density of 0.4wb/m²

The total m.m.f is given by,

$\text{m.m.f} = H_{ring} l_{ring} +H_{gap} l_{gap}$

$\Rightarrow \text{m.m.f} = \frac{B}{\mu_r \mu_0} l_{ring} +\frac{B}{ \mu_0} l_{gap}$

Substituting the given values in the above expression to get;

$\Rightarrow \text{m.m.f} = \frac{0.4 \times 60 \times 10^{-2} }{400 \times 4 \pi \times 10^{-7}} + \frac{0.4 \times 6 \times 10^{-3} }{4 \pi \times 10^{-7}} =795.77 \text{A.T.}$

Hence, 795.77 A.T. is the m.m.f. required to produce a flux density of 0.4wb/m²