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
Expert's answer
2011-06-10T05:32:24-0400
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/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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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
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Dear visitor, please use panel for submitting new questions
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
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