A long straight wire of radius a carries a steady current I. The current is uniformly distribute over its cross-section. The ratio of the magnetic fields B and B′, at radial distances (a/2) and 2a respectively, from the axis of the wire is:
1
4
12
14
A
14
B
1
C
4
D
12
Open in App
Solution
Verified by Toppr
Let J be the surface current density . So, I=J.πa2
Using Ampere's law the field at radial distance a/2: B.2π(a/2)=μ0Ien
Here Ien=current enclosed by amperical loop of radius a/2 =J.π(a/2)2=I/4
So, B=μ0I4πa
And at radial distance 2a: B′.2π(2a)=μ0Ien
Here Ien=I as it is outside the wire.
Thus, B′=μ0I4πa
∴BB′=1
Was this answer helpful?
1
Similar Questions
Q1
A long straight wire of radius a carries a steady current I. The current is uniformly distribute over its cross-section. The ratio of the magnetic fields B and B′, at radial distances (a/2) and 2a respectively, from the axis of the wire is:
View Solution
Q2
A long straight wire of radius a carries a steady current i. The current is uniformly distributed across its cross section. The ratio of the magnetic field at a/2 and 2a is:
View Solution
Q3
A long, straight wire, of radius a, carries a current distributed uniformly over its cross-section. The ratio of the magnetic fields due to the wire, at distances a3and 2a, respectively from the axis of the wire, is: