Skip to the question

JEE MainPhysics

Reviewed by official_key

+4 marks−1 if incorrectSingle correctpyq

Magnetic Effects of Current and Magnetism: Physics | JEE Main

A long straight wire of circular cross-section (radius a) is carrying steady current I. The current I is uniformly distributed across this cross-section. The magnetic field is

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

AZero in the region

\mathrm{r} < \mathrm{a}

and inversely proportional to

\mathrm{r}

in the region

\mathrm{r} > \mathrm{a}

BInversely proportional to

\mathrm{r}

in the region

\mathrm{r} < \mathrm{a}

and uniform throughout in the region

\mathrm{r} > \mathrm{a}

CDirectly proportional to

\mathrm{r}

in the region

\mathrm{r} < \mathrm{a}

and inversely proportional to

\mathrm{r}

in the region

\mathrm{r} > \mathrm{a}

DUniform in the region

\mathrm{r} < \mathrm{a}

and inversely proportional to distance

\mathrm{r}

from the axis, in the region

\mathrm{r} > \mathrm{a}

Step-by-step solutionView

Correct answer

By Ampère's circuital law, the magnetic field is directly proportional to $r$ inside the wire ( $r < a$ ) and inversely proportional to $r$ outside the wire ( $r > a$ ).

Option analysis

Why each option works or fails

A · Zero in the region

\mathrm{r} < \mathrm{a}

and inversely proportional to

\mathrm{r}

in the region

\mathrm{r} > \mathrm{a}

Confusing a solid wire carrying uniform volume current with a hollow cylindrical shell carrying only surface current. Recognize that a solid wire has current distributed throughout its cross-section, so an Ampèrian loop of radius $r < a$ encloses a nonzero current proportional to $r^2$ .

B · Inversely proportional to

\mathrm{r}

in the region

\mathrm{r} < \mathrm{a}

and uniform throughout in the region

\mathrm{r} > \mathrm{a}

Applying the $1/r$ dependence inside the conductor and assuming the field reaches a constant value outside. Apply Ampère's law separately: inside, $I_{\text{enc}} \propto r^2$ , leading to $B \propto r$ ; outside, $I_{\text{enc}} = I$ , leading to $B \propto 1/r$ .

C · Directly proportional to

\mathrm{r}

in the region

\mathrm{r} < \mathrm{a}

and inversely proportional to

\mathrm{r}

in the region

\mathrm{r} > \mathrm{a}

None. This option correctly applies Ampère's circuital law to both regions. Inside the wire, $B(2\pi r) = \mu_0 I (r^2/a^2) \implies B \propto r$ . Outside the wire, $B(2\pi r) = \mu_0 I \implies B \propto 1/r$ .

D · Uniform in the region

\mathrm{r} < \mathrm{a}

and inversely proportional to distance

\mathrm{r}

from the axis, in the region

\mathrm{r} > \mathrm{a}

Assuming the magnetic field inside a current-carrying cylinder is uniform, similar to the field inside an ideal solenoid. A straight wire generates concentric circular magnetic field lines where the enclosed current increases with radius, giving a field that increases linearly with $r$ , not a uniform field.

Reviewed route

Solution

StepWorking

01concept

Apply Ampère's circuital law: $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}}$ . By symmetry, $B(2\pi r) = \mu_0 I_{\text{enc}}$ . For $r < a$ , current enclosed is $I_{\text{enc}} = I \frac{\pi r^2}{\pi a^2} = I \frac{r^2}{a^2}$ , giving $B = \frac{\mu_0 I r}{2\pi a^2} \propto r$ . For $r > a$ , total current is enclosed $I_{\text{enc}} = I$ , giving $B = \frac{\mu_0 I}{2\pi r} \propto \frac{1}{r}$ .

02option_verdict

Incorrect because field inside is zero only for a hollow thin-walled cylinder, not a solid uniform current-carrying wire.

03option_verdict

Incorrect because inside the wire $B$ scales linearly with $r$ , and outside the wire $B$ decays as $1/r$ rather than remaining uniform.

04option_verdict

Correct. $B \propto r$ for $r < a$ due to quadratic area scaling of enclosed current, and $B \propto 1/r$ for $r > a$ as the total enclosed current remains constant.

05option_verdict

Incorrect because current density is uniform, which makes the enclosed current grow as $r^2$ , so $B$ increases linearly with $r$ rather than staying uniform.

✓discriminator

Inside: $I_{\text{enc}} \propto r^2 \implies B(2\pi r) \propto r^2 \implies B \propto r$ . Outside: $I_{\text{enc}} = \text{const} \implies B \propto 1/r$ .

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Physics
Concepts assessed: Physics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 7 September 2026

Quick checks

Students also ask

Why does current enclosed scale with

r^2

for

r < a

?

Because current is uniformly distributed across the cross-sectional area, current density $J = I/(\pi a^2)$ is constant, so $I_{\text{enc}} = J \cdot (\pi r^2) = I (r/a)^2$ .

Answer

By Ampère's circuital law, the magnetic field is directly proportional to $r$ inside the wire ( $r < a$ ) and inversely proportional to $r$ outside the wire ( $r > a$ ).

Why each option works or fails

A: Zero in the region $\mathrm{r} < \mathrm{a}$ and inversely proportional to $\mathrm{r}$ in the region $\mathrm{r} > \mathrm{a}$ - Confusing a solid wire carrying uniform volume current with a hollow cylindrical shell carrying only surface current. Recognize that a solid wire has current distributed throughout its cross-section, so an Ampèrian loop of radius $r < a$ encloses a nonzero current proportional to $r^2$ .
B: Inversely proportional to $\mathrm{r}$ in the region $\mathrm{r} < \mathrm{a}$ and uniform throughout in the region $\mathrm{r} > \mathrm{a}$ - Applying the $1/r$ dependence inside the conductor and assuming the field reaches a constant value outside. Apply Ampère's law separately: inside, $I_{\text{enc}} \propto r^2$ , leading to $B \propto r$ ; outside, $I_{\text{enc}} = I$ , leading to $B \propto 1/r$ .
C · correct: Directly proportional to $\mathrm{r}$ in the region $\mathrm{r} < \mathrm{a}$ and inversely proportional to $\mathrm{r}$ in the region $\mathrm{r} > \mathrm{a}$ - None. This option correctly applies Ampère's circuital law to both regions. Inside the wire, $B(2\pi r) = \mu_0 I (r^2/a^2) \implies B \propto r$ . Outside the wire, $B(2\pi r) = \mu_0 I \implies B \propto 1/r$ .
D: Uniform in the region $\mathrm{r} < \mathrm{a}$ and inversely proportional to distance $\mathrm{r}$ from the axis, in the region $\mathrm{r} > \mathrm{a}$ - Assuming the magnetic field inside a current-carrying cylinder is uniform, similar to the field inside an ideal solenoid. A straight wire generates concentric circular magnetic field lines where the enclosed current increases with radius, giving a field that increases linearly with $r$ , not a uniform field.

Step-by-step solution

concept: Apply Ampère's circuital law: $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}}$ . By symmetry, $B(2\pi r) = \mu_0 I_{\text{enc}}$ . For $r < a$ , current enclosed is $I_{\text{enc}} = I \frac{\pi r^2}{\pi a^2} = I \frac{r^2}{a^2}$ , giving $B = \frac{\mu_0 I r}{2\pi a^2} \propto r$ . For $r > a$ , total current is enclosed $I_{\text{enc}} = I$ , giving $B = \frac{\mu_0 I}{2\pi r} \propto \frac{1}{r}$ .
option_verdict: Incorrect because field inside is zero only for a hollow thin-walled cylinder, not a solid uniform current-carrying wire.
option_verdict: Incorrect because inside the wire $B$ scales linearly with $r$ , and outside the wire $B$ decays as $1/r$ rather than remaining uniform.
option_verdict: Correct. $B \propto r$ for $r < a$ due to quadratic area scaling of enclosed current, and $B \propto 1/r$ for $r > a$ as the total enclosed current remains constant.
option_verdict: Incorrect because current density is uniform, which makes the enclosed current grow as $r^2$ , so $B$ increases linearly with $r$ rather than staying uniform.
discriminator: Inside: $I_{\text{enc}} \propto r^2 \implies B(2\pi r) \propto r^2 \implies B \propto r$ . Outside: $I_{\text{enc}} = \text{const} \implies B \propto 1/r$ .