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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.

one year ago

Answers : (2)

Arun
25089 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
one year ago
Vikas TU
13786 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  
 
one year ago
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