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Grade: 12
        
A verical circular coil of radius 0.1 m and of 10 turns carries a steady current  .When the plane of the coil is normal  to the magnetic meridian, a neutral  point  is observed at the centre of the coil.if    BH=horizontal component of earth's magnetic field =0.314×10^-4 T then find the current  flowing  through the coil.
4 months ago

Answers : (2)

Arun
23769 Points
							
Dear student
As we know
B = mu * N* i /2r
hence
I = 2 B* r / 2 N* mu
Now put the value and you will get the amswer 0.5 pi
Hope it helps
4 months ago
Vikas TU
10470 Points
							
Dear student 
At neutral point 
Magnetic field become opposite and equal to some other magnetic field , 
Other magnetic field = Horizontal component of earth magnetic field.
Bh = magnetic field at centre of crcular coil = (μ0/4π)*2pie nI /r 
I = (μ0/4π)* rBh /2pie*n 
(10^7*0.1*0.314*10^-4)/ (2*3.14*10) = 0.5A 
Hope this helps  
 
4 months ago
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