The magnetic field intensity outside the coaxial cable is
H=0
H = NI/d
H = 1 p/2 pi a^2
H = 1//2 ppi
The Correct Answer and Explanation is :
Correct Answer:
The magnetic field intensity outside a coaxial cable is H = 0.
Explanation:
A coaxial cable consists of two concentric cylindrical conductors: an inner conductor carrying current II and an outer conductor that carries a return current of equal magnitude but opposite direction. The key to understanding why the magnetic field intensity outside the coaxial cable is zero lies in Ampère’s Law and the principle of superposition.
1. Applying Ampère’s Circuital Law
Ampère’s Circuital Law states: ∮H⋅dl=Ienc\oint \mathbf{H} \cdot d\mathbf{l} = I_{\text{enc}}
where IencI_{\text{enc}} is the total current enclosed by an Amperian loop.
- Inside the inner conductor: The current is distributed over the cross-section.
- Between the inner and outer conductors: The field depends on enclosed current.
- Outside the coaxial cable: The inner conductor carries a current II in one direction, and the outer conductor carries −I in the opposite direction. If we draw an Amperian loop outside the cable (with a radius greater than the outer conductor), the total enclosed current is: Ienc=I−I=0I_{\text{enc}} = I – I = 0 Since no net current is enclosed, Ampère’s Law results in H=0H = 0 outside.
2. Superposition of Magnetic Fields
Each conductor generates a magnetic field according to Biot-Savart’s Law. The field from the inner conductor outside the cable is canceled exactly by the field from the outer conductor due to the opposite direction of current. This perfect cancellation ensures that the net magnetic field outside the coaxial cable is zero.
Conclusion:
Since the net current enclosed by any Amperian loop outside the coaxial cable is zero, the magnetic field intensity in this region must be H = 0.