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# 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
2011-06-10T05:32:24-0400
To find the flux of field there is a formula
B=&mu; *N *I/(2*&pi;*r)

r- the mean radius. if the mean circumference s=0.6 m, the the mean radius
r=s/(2*&pi;)=0.1
the value of munot is constant: &mu;=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/m2 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/m2

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05.11.18, 22:59

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Jha Devarsh
05.11.18, 16:55

2 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