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Grade 12Magnetism

plz help a circular current carrying coil has radius r. Then distance from which centre of coil on the axis when the magnetic inductions will be 1/8 of its value at centre of coil is...?

Profile image of harshit goswami
15 Years agoGrade 12
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2 Answers

Profile image of vikas askiitian expert
15 Years ago

B(axis) = uoIr2/2(r2+x2)3/2

B(center) = uoI/2r

 

B(axis) = 1/8 (B(center) so

r2/(r2+x2)3/2 = 1/8r

8r3 = (r2+x2)3/2

(2r)2  =  r2 + x2

 x = r(root3)

 

approve if u like my ans

Profile image of Kushagra Madhukar
5 Years ago
Dear student,
Please find the attached solution to your problem
B(axis) = uoIr2/2(r2+x2)3/2
B(center) = uoI/2r
B(axis) = 1/8 * B(center)
so
r2/(r2+x2)3/2 = 1/8r
8r3 = (r2+x2)3/2
(2r)2 = r2 + x2
x = r√3
Thanks and regards,
Kushagra

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