In a co-axial straight cable, the central conductor and the outer conductor carry equal currents in opposite directions. The magnetic field is zero :

Magnetics PYQ — JEE Main 2024 · 5 Apr · Shift 1 · Q79 with Solution

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Correct answer: B

Use Ampere’s law / enclosed current idea

For a long coaxial cable, the magnetic field at distance $r$ from the axis depends on the net current enclosed:

B(2\pi r)=\mu_0 I_{\text{enclosed}}

So,

B=\frac{\mu_0 I_{\text{enclosed}}}{2\pi r}

The central conductor carries current $+I$ , and the outer conductor carries current $-I$ of equal magnitude.

Check different regions

(i) Outside the cable

If the Amperian loop is taken outside the outer conductor, it encloses both currents:

I_{\text{enclosed}} = I + (-I) = 0

Hence,

B=0

So the magnetic field is zero outside the cable.

(ii) Between the two conductors

Here the loop encloses only the inner conductor current $I$ .

Thus,

I_{\text{enclosed}}=I Rightarrow B=\frac{\mu_0 I}{2\pi r}\neq 0

So it is not zero between the conductors.

(iii) Inside the outer conductor

Within the material of the outer conductor, only part of the return current is enclosed, so in general the enclosed current is not necessarily zero at every point. Thus the magnetic field is generally not zero throughout this region.

(iv) Inside the inner conductor

Inside the inner conductor, enclosed current depends on radius (assuming uniform current density), so

I_{\text{enclosed}} \propto r^2

Hence $B\neq 0$ in general (except at the center $r=0$ only).

Evaluate options

A: inside the outer conductor — Not generally zero.
B: outside the cable — Correct, because net enclosed current is zero.
C: in between the two conductors — Not zero.
D: inside the inner conductor — Not generally zero.

Final answer

The magnetic field is zero outside the cable.

\boxed{\text{B}}

More from Magnetics

View written solutionFree

Correct answer: B

Use Ampere’s law / enclosed current idea

For a long coaxial cable, the magnetic field at distance $r$ from the axis depends on the net current enclosed:

B(2\pi r)=\mu_0 I_{\text{enclosed}}

So,

B=\frac{\mu_0 I_{\text{enclosed}}}{2\pi r}

The central conductor carries current $+I$ , and the outer conductor carries current $-I$ of equal magnitude.

Check different regions

(i) Outside the cable

If the Amperian loop is taken outside the outer conductor, it encloses both currents:

I_{\text{enclosed}} = I + (-I) = 0

Hence,

B=0

So the magnetic field is zero outside the cable.

(ii) Between the two conductors

Here the loop encloses only the inner conductor current $I$ .

Thus,

I_{\text{enclosed}}=I Rightarrow B=\frac{\mu_0 I}{2\pi r}\neq 0

So it is not zero between the conductors.

(iii) Inside the outer conductor

(iv) Inside the inner conductor

Inside the inner conductor, enclosed current depends on radius (assuming uniform current density), so

I_{\text{enclosed}} \propto r^2

Hence $B\neq 0$ in general (except at the center $r=0$ only).

Evaluate options

A: inside the outer conductor — Not generally zero.
B: outside the cable — Correct, because net enclosed current is zero.
C: in between the two conductors — Not zero.
D: inside the inner conductor — Not generally zero.

Final answer

The magnetic field is zero outside the cable.

\boxed{\text{B}}

Magnetics question

Magnetics question

(i) Outside the cable

(ii) Between the two conductors

(iii) Inside the outer conductor

(iv) Inside the inner conductor

More from Magnetics

Magnetics question

Magnetics question

(i) Outside the cable

(ii) Between the two conductors

(iii) Inside the outer conductor

(iv) Inside the inner conductor

More from Magnetics