Question #3063

A coil 0f 200 turns is wound uniformly over wooden ring having a mean circum ference of 600mm and a uniformly cross the sectional area of 500 mm^2 IF the current through the coil is 4A calculate

1)the magnetic field strength

2)the flux density

3)the total flux

1)the magnetic field strength

2)the flux density

3)the total flux

Expert's answer

To find the flux of field there is a formula

**B=μ *N *I/(2*π*r)**

r- the mean radius. if the mean circumference s=0.6 m, the the mean radius

**r=s/(2*****π****)=0.1**

the value of munot is constant:**μ**=4(3.142)x10^{-7} web/amp m

Hence we have

B=4*3.142x10^{-7} * 200*4/(2*3.142*0.1)

B=1.6x10^{-3} weber/m^{2} or tesla

So we get magnetic field

For finding the magnetic flux density there is a formula

**magnetic flux=magnetic field * Area**

magnetic flux=1.6x10^{-3} * 0.5x10^{-3}

magnetic flux=0.8x10^{-6} webers

Now flux density

**flux density=magnetic flux/area**

**flux density=1.6x10**^{-3} web/m^{2}

r- the mean radius. if the mean circumference s=0.6 m, the the mean radius

the value of munot is constant:

Hence we have

B=4*3.142x10

B=1.6x10

So we get magnetic field

For finding the magnetic flux density there is a formula

magnetic flux=1.6x10

magnetic flux=0.8x10

Now flux density

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## Comments

Assignment Expert05.11.18, 22:59Dear visitor, please use panel for submitting new questions

Jha Devarsh05.11.18, 16:552 mWb is to be produced in the air gap of the magnetic circuit shown in figure. How much ampere turns the coil must provide to achieve this? Relative permeability μr of the core material may be assumed to be constant and equal to 5000. All the dimensions shown are in cm and the sectional area is 25cm2 throughout

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