Q.A horizontal overhead power line carries a current of $90\ \text{A}$ in east to west direction. What is the magnitude and direction of the magnetic field due to the current $1.5\ \text{m}$ below the line?

Showing the 12 most recent of 16 on this concept.

1. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.A wire carrying current I and other parallel wire carrying current 2I in the same direction produces a magnetic field B at the midpoint between them. Then the magnitude of field at the same point, when the 2I wire is switched off (A) B/2 (B) 2 B (C) B (D) 4 B

   ›Reveal solutionSolution

   Because the two wires' fields oppose at the midpoint, the net field with both wires on already equals the lone-$I$-wire's field; switching off the $2I$ wire therefore leaves the field magnitude unchanged at B.

   Concept and Intuition

   For two infinite parallel wires carrying current in the same direction, the magnetic field circles each wire by the right-hand rule. At a point exactly between them, the two contributions point in opposite directions (one wire's field curls one way past the midpoint, the other's curls the opposite way past the midpoint) — they don't add, they subtract. This is different from the case of currents in opposite directions, where the midpoint fields add. Here, since both wires are equidistant from the midpoint, the stronger $2I$ wire's field "wins" the subtraction, and the net field magnitude equals the difference, which numerically works out to be exactly equal to the field of the lone $I$ wire.

   Step-by-Step Solution

   1. Let the wires be separated by distance $d$; the midpoint is at distance $d/2$ from each.
   2. Field due to $I$ alone at the midpoint: $B_I = \dfrac{\mu_0 I}{2\pi (d/2)} = \dfrac{\mu_0 I}{\pi d}$.
   3. Field due to $2I$ alone at the midpoint: $B_{2I} = \dfrac{\mu_0 (2I)}{\pi d} = 2B_I$.
   4. Since both currents flow the same way, these two fields point in opposite directions at the midpoint, so the net field with both wires on is $B = B_{2I} - B_I = 2B_I - B_I = B_I$. …

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2. AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
   Q.If a straight infinitely long horizontal wire carries a current of 50 A in east-west direction, then the magnitude of the magnetic field due to the current at a vertical distance of 2 m above the wire is (A) 10 μT (B) 2.5 μT (C) 5 μT (D) 7.5 μT

   ›Reveal solutionSolution

   A direct application of the formula for the magnetic field due to a long straight current-carrying wire at a perpendicular distance.

   Concept and Intuition

   Ampere's law gives the field around an infinitely long straight wire as circles centred on the wire, with magnitude falling off as $1/d$ from the wire — independent of direction (east-west orientation doesn't change the magnitude, only which way the field circles point).

   Step-by-Step Solution

   1. Formula: $B=\dfrac{\mu_0 I}{2\pi d}$, with $\mu_0/2\pi = 2\times10^{-7}\,T\cdot m/A$. …

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3. AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQ
   Q.Magnetic field at 0.1m from a long straight wire carrying 10A current is (A) $2 \times 10^{-5}$ T (B) $2 \times 10^{-4}$ T (C) $2 \times 10^{-6}$ T (D) $10^{-5}$ T

   ›Reveal solutionSolution

   Direct application of the straight-wire magnetic field formula $B=\mu_0 I/(2\pi r)$ gives $2\times10^{-5}$ T.

   Concept and Intuition

   A long current-carrying straight wire produces a magnetic field that circles around it, with magnitude falling off as $1/r$ from the wire — this follows from Ampere's circuital law applied to a circular loop of radius $r$ around the wire.

   Step-by-Step Solution

   1. Formula: $B=\dfrac{\mu_0 I}{2\pi r}$, with $\mu_0=4\pi\times10^{-7}\ \text{T m/A}$.
   2. Substitute $I=10$ A, $r=0.1$ m: …

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4. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.The magnetic field at a distance of 10 cm from a long straight thin wire carrying a current of 4 A is (A) $6\,\mu\mathrm{T}$ (B) $16\,\mu\mathrm{T}$ (C) $8\,\mu\mathrm{T}$ (D) $4\,\mu\mathrm{T}$

   ›Reveal solutionSolution

   This tests the magnetic field due to a long straight current-carrying wire (Ampere's law / Biot-Savart result). Answer: $8\ \mu\mathrm{T}$.

   Concept and Intuition

   A long straight wire carrying current $I$ produces a magnetic field that circles around the wire, with magnitude falling off as $1/r$ from the wire (unlike a point charge's field, which falls as $1/r^2$), because the field is due to a continuous line of current rather than a point source. This comes directly from Ampere's circuital law applied to a circular loop around the wire.

   Step-by-Step Solution

   1. Formula: $B=\dfrac{\mu_0 I}{2\pi r}$, with $\mu_0=4\pi\times10^{-7}\ \text{T m/A}$.
   2. Substitute $I=4$ A, $r=10\text{ cm}=0.1$ m: $B=\dfrac{4\pi\times10^{-7}\times4}{2\pi\times0.1}$.
   3. The $\pi$ cancels: $B=\dfrac{4\times10^{-7}\times4}{2\times0.1}=\dfrac{16\times10^{-7}}{0.2}=8\times10^{-6}$ T. …

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5. AP EAPCET 2025Set ap-2025-05-19-FN1 markMCQ
   Q.The magnetic field at a point P at a distance of $2\ \mathrm{cm}$ from a long straight wire of diameter $0.5\ \mathrm{mm}$ carrying a current of $1\ \mathrm{A}$ is B. If the diameter of the wire is doubled without changing the current, the magnetic field at the same point P is (A) 2B (B) $\dfrac{B}{2}$ (C) $\dfrac{3B}{4}$ (D) B

   ›Reveal solutionSolution

   Outside a straight current-carrying wire, the magnetic field depends only on the enclosed current and the distance from the wire's axis (Ampere's law), not on the wire's thickness. Doubling the wire's diameter (while it stays much thinner than the 2 cm distance to P) leaves the external field unchanged.

   Concept and Intuition

   By Ampère's Circuital Law, for a point outside a straight wire, $\oint \vec B \cdot d\vec l = \mu_0 I_{enc}$, and by symmetry this gives $B = \dfrac{\mu_0 I}{2\pi d}$ where $d$ is the perpendicular distance from the wire's axis to the field point. This result is independent of how the current is distributed within the wire's cross-section, as long as the field point lies outside the wire. So changing the wire's diameter (while P remains outside the wire, i.e. $d$ measured from the same axis) does not affect $B$ at P, provided the current $I$ is unchanged.

   Step-by-Step Solution

   1. Point P is at $d = 2\,\mathrm{cm} = 20\,\mathrm{mm}$ from the wire's axis.
   2. The wire's radius is $0.25\,\mathrm{mm}$ initially and $0.5\,\mathrm{mm}$ after doubling the diameter — in both cases far smaller than $20\,\mathrm{mm}$, so P remains well outside the wire. …

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6. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
   Q.The maximum magnetic field produced by a current of 12 A passing through a copper wire of diameter 1.2 mm is (A) 2 mT (B) 4 mT (C) 1.5 mT (D) 8 mT

   ›Reveal solutionSolution

   This tests the field of a long straight current-carrying wire evaluated at its own surface (where it is maximum), giving $B=4$ mT for a 12 A current in a 1.2 mm diameter wire.

   Concept and Intuition

   For a straight current-carrying wire (treated as a long cylindrical conductor), the magnetic field at a perpendicular distance $r$ from the axis grows linearly with $r$ inside the wire (uniform current density) and falls off as $1/r$ outside it. The field is therefore maximum exactly at the wire's surface, where $r$ equals the wire's radius $a$: $B_{max} = \dfrac{\mu_0 I}{2\pi a}$.

   Step-by-Step Solution

   1. Diameter $= 1.2\,\mathrm{mm}$, so radius $a = 0.6\,\mathrm{mm} = 6\times10^{-4}\,\mathrm{m}$.
   2. Use $B = \dfrac{\mu_0 I}{2\pi a} = \left(\dfrac{\mu_0}{2\pi}\right)\dfrac{I}{a}$, and $\dfrac{\mu_0}{2\pi} = 2\times10^{-7}\,\mathrm{T\,m\,A^{-1}}$. …

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7. AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
   Q.A wire shaped in a regular hexagon of side 2 cm carries a current of 4 A. The magnetic field at the centre of hexagon is. [FIGURE] (a regular hexagon with vertices labelled a at bottom-left, b at left, c at top-left, d at top-right, e at right, f at bottom-right, and the centre marked O) (A) $4\sqrt{3} \times 10^{-5}$ T (B) $8\sqrt{3} \times 10^{-5}$ T (C) $\sqrt{3} \times 10^{-5}$ T (D) $6\sqrt{3} \times 10^{-5}$ T

   ›Reveal solutionSolution

   This tests the magnetic field at the centre of a regular current-carrying polygon, built up side by side using the finite-straight-wire (Biot–Savart) formula.

   Concept and Intuition

   Each side of the hexagon is a finite straight current segment. The field it produces at the centre can be found from the standard finite-wire formula, using the perpendicular distance from the centre to that side (the "apothem") and the half-angle each side subtends at the centre. By symmetry all six sides contribute equally, so the total field is six times one side's contribution.

   Step-by-Step Solution

   1. For a finite straight wire, field at perpendicular distance $d$ subtending half-angles $\theta$ at each end: $B_{side}=\dfrac{\mu_0 I}{4\pi d}(\sin\theta+\sin\theta)=\dfrac{\mu_0 I\sin\theta}{2\pi d}$.
   2. For a regular hexagon ($n=6$ sides), each side subtends a half-angle $\theta=\pi/n=30^\circ$ at the centre, and the apothem is $d=\dfrac{a}{2}\cot(\pi/n)=\dfrac{a}{2}\cot30^\circ=\dfrac{a\sqrt3}{2}$.
   3. Total field from all 6 sides simplifies to the standard result $B=\dfrac{n\mu_0 I}{2\pi a}\sin(\pi/n)\tan(\pi/n)$; for $n=6$ this reduces neatly to $B=\dfrac{\sqrt3\,\mu_0 I}{\pi a}$. …

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8. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
   Q.Two long straight parallel conductors A and B carrying currents 4.5 A and 8 A respectively are separated by 25 cm in air. The resultant magnetic field at a point which is at a distance of 15 cm from conductor A and 20 cm from conductor B is (A) $2 \times 10^{-5}\ \mathrm{N}$ (B) $2 \times 10^{-4}\ \mathrm{N}$ (C) $10^{-5}\ \mathrm{N}$ (D) $10^{-4}\ \mathrm{N}$

   ›Reveal solutionSolution

   The distances 15, 20, 25 cm form a right triangle, which makes the two wires' fields at that point mutually perpendicular; their Pythagorean sum comes out to $1\times10^{-5}\ \text{T}$.

   Concept and Intuition

   The field from a long straight wire circles around it, always perpendicular to the line joining the wire to the field point. So if the lines from the two wires to our point happen to be perpendicular to each other (as they are here, since $15$–$20$–$25$ is a Pythagorean triple, meaning the angle at the point is $90°$), the two field vectors (each perpendicular to its own line) are perpendicular to each other too. That lets us combine them with the Pythagorean theorem instead of a general cosine-rule addition.

   Step-by-Step Solution

   1. Check the geometry: $15^2+20^2 = 225+400=625=25^2$. So the triangle formed by the point and the two wires has a right angle at the point (between the lines to wire $A$ and wire $B$).
   2. Field due to wire $A$ ($I_1=4.5\ \text{A}$, $r_1=15\ \text{cm}=0.15\ \text{m}$): $B_1 = \dfrac{\mu_0 I_1}{2\pi r_1} = \dfrac{2\times10^{-7}\times4.5}{0.15}=6\times10^{-6}\ \text{T}$.
   3. Field due to wire $B$ ($I_2=8\ \text{A}$, $r_2=20\ \text{cm}=0.20\ \text{m}$): $B_2=\dfrac{\mu_0 I_2}{2\pi r_2}=\dfrac{2\times10^{-7}\times8}{0.20}=8\times10^{-6}\ \text{T}$. …

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9. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
   Q.Two infinite length wires carry currents 8 A and 6 A respectively and are placed along X and Y axes respectively. Magnetic field at a point P (0, 0, d) will be (A) $\dfrac{7\mu_0}{\pi d}$ (B) $\dfrac{10\mu_0}{\pi d}$ (C) $\dfrac{14\mu_0}{\pi d}$ (D) $\dfrac{5\mu_0}{\pi d}$

   ›Reveal solutionSolution

   The point P is equidistant from both wires, and the two fields there point along perpendicular directions, so they combine via Pythagoras (the classic 8-6-10 triple).

   Concept and Intuition

   For an infinite straight wire, the field magnitude at perpendicular distance $s$ is $B=\mu_0 I/2\pi s$, directed tangentially around the wire (right-hand rule). Here, point $P(0,0,d)$ lies on the z-axis, at perpendicular distance $d$ from both the x-axis wire and the y-axis wire. Because the two wires are along orthogonal axes, the field contributions at P (each tangential to its own wire, i.e. lying in the plane perpendicular to that wire) end up pointing along mutually perpendicular directions ($\hat x$ and $\hat y$ roughly), so we must add them as vectors rather than algebraically.

   Step-by-Step Solution

   1. Perpendicular distance from P$(0,0,d)$ to the x-axis (wire along x, through origin) is $d$; likewise the perpendicular distance to the y-axis is $d$.
   2. Field magnitude due to the 8 A wire: $B_1 = \dfrac{\mu_0(8)}{2\pi d}$.
   3. Field magnitude due to the 6 A wire: $B_2 = \dfrac{\mu_0(6)}{2\pi d}$. …

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10. AP EAPCET 2023Set eng-2023-05-18-AN1 markMCQ
    Q.A straight wire carrying a current of 12 A is bent into a semi-circular arc of radius 2 cm as shown in the figure. Then the magnetic field due to the straight segments at the centre of the arc is [FIGURE] (a straight wire carrying current 12 A, bent into a semicircular arc of radius 2 cm; point O marks the centre of the arc) (A) 12 T (B) 6 T (C) 24 T (D) 0

    ›Reveal solutionSolution

    This tests recognizing when the Biot-Savart contribution from a straight wire vanishes at a field point. Since O lies on the same line as the straight segments, their combined field there is exactly zero.

    Concept and Intuition

    The Biot-Savart law gives the field due to a current element as $d\vec B = \dfrac{\mu_0}{4\pi}\dfrac{I\,d\vec l\times\hat r}{r^2}$, where $\hat r$ points from the element to the field point. If the field point lies exactly on the line containing the straight wire (whether ahead of it, behind it, or on it), then $d\vec l$ and $\hat r$ are always parallel or antiparallel for every element of that straight segment, making the cross product zero everywhere along the segment. Here, point O lies on the baseline that both straight segments lie along, so this condition is satisfied for both.

    Step-by-Step Solution

    1. Identify the geometry: the straight parts of the wire lie along a single baseline, and O sits on that same baseline (directly below the semicircular bulge, at the arc's centre).
    2. For any element $d\vec l$ of a straight segment lying along this baseline, the vector $\hat r$ from that element to O is also along the baseline (since O is on the line).
    3. Therefore $d\vec l \times \hat r = 0$ for every element of both straight segments. …

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11. AP EAPCET 2022Set ap-2022-07-12-FN1 markMCQ
    Q.A current of 5A exists in a square loop of side length 5 cm. The magnitude of magnetic field at the centre of the square loop is approximately. (A) 3.8 mT (B) 1.7 mT (C) 4.8 mT (D) 2.3 mT

    ›Reveal solutionSolution

    For a square loop, $B_{centre} = \dfrac{2\sqrt{2}\,\mu_0 I}{\pi a}$. The stated data give $\approx 0.11\ mT$, which is about $20\times$ below every option, so the printed choices appear misprinted; per the official key the marked answer is $2.3\ mT$.

    Formula. The field at the centre of a square loop of side $a$ carrying current $I$ is

    $B = \frac{2\sqrt{2}\,\mu_0 I}{\pi a},$

    obtained by adding the four sides, each contributing $\dfrac{\mu_0 I}{4\pi (a/2)}\,(2\sin 45^\circ)$.

    Substitute $I = 5\ A$, $a = 5\ cm = 0.05\ m$:

    $B = \frac{2\sqrt{2}\,(4\pi\times10^{-7})(5)}{\pi (0.05)} = 2\sqrt{2}\times 4\times10^{-7}\times 100 = 1.13\times10^{-4}\ T \approx 0.11\ mT.$ …

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12. AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
    Q.A long current carrying wire produces a magnetic field of 1 T at a distance of r. The magnetic field at

    (a) $\dfrac{r}{2}$

    (b) 2r and

    (c) 3r is (A)

    (a) 2T,

    (b) = $\dfrac{1}{2}$ T,

    (c) = $\dfrac{1}{3}$ T (B)

    (a) 3T,

    (b) = $\dfrac{1}{3}$ T,

    (c) = $\dfrac{1}{6}$ T (C)

    (a) $\dfrac{3}{2}$ T,

    (b) = $\dfrac{1}{4}$ T,

    (c) = $\dfrac{1}{8}$ T (D)

    (a) $\dfrac{5}{2}$ T,

    (b) = $\dfrac{1}{2}$ T,

    (c) = $\dfrac{1}{3}$ T

    ›Reveal solutionSolution

    The field of a long straight wire falls off as $1/r$ (unlike a point charge's $1/r^2$). Halving the distance doubles the field; at $2r$ and $3r$ the field is $1/2$ and $1/3$ of the original.

    Concept and Intuition

    From Ampère's law, $B = \dfrac{\mu_0 I}{2\pi r}$ for an infinite straight wire — a simple inverse relationship with distance, since the field lines are concentric circles whose "density" thins out linearly (not quadratically like a point source) as you move away from a line source.

    Step-by-Step Solution

    1. Given $B(r) = 1$ T, and $B \propto 1/r$, write $B(r) \cdot r = \text{constant} = 1\times r$.
    2. At $r/2$: $B\times(r/2) = 1\times r \Rightarrow B = 2$ T.
    3. At $2r$: $B\times(2r) = 1\times r \Rightarrow B = \dfrac12$ T. …

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