# Answer to Question #3063 in Electromagnetism for asimomar

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

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

**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 mHence we have

B=4*3.142x10

^{-7}* 200*4/(2*3.142*0.1)B=1.6x10

^{-3}weber/m^{2}or teslaSo 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}webersNow flux density

**flux density=magnetic flux/area****flux density=1.6x10**^{-3}web/m^{2}Need a fast expert's response?

Submit orderand get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

## Comments

## Leave a comment