Q.Consider a tightly wound 200 turns circular coil of radius 20 cm carrying a current of 1A. What is the magnitude of the magnetic field at the centre of the coil?

Showing the 12 most recent of 14 on this concept.

1. CBSE 2026Set ANNUAL1 markMCQ
   Q.An arc of a circle of radius R subtends an angle π/2 at the centre. It carries a current i. The magnetic induction at the centre will be

   (a) μ0 i / 2R

   (b) μ0 i / 8R

   (c) μ0 i / 4R

   (d) 2μ0 i / 5R

   ›Reveal solutionSolution

   The magnetic induction at the centre due to a current-carrying arc subtending angle $\theta$ is $\dfrac{\mu_0 i}{4\pi R}\theta$; for $\theta=\pi/2$ this gives $\mu_0 i/(8R)$.

   For a full circular loop of radius $R$ carrying current $i$, the magnetic field at the centre is $B_{\text{full}} = \dfrac{\mu_0 i}{2R}$, using the Biot–Savart law integrated over the full $2\pi$ angle.

   …

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2. CBSE 2025Set 55/6/11 markMCQ
   Q.The value of magnetic field at point O in the given figure is: (A) $\dfrac{\mu_0 I}{2\pi R}$ (B) $\dfrac{\mu_0 I}{\pi R}$ (C) $\dfrac{\mu_0 I}{4R}$ (D) $\dfrac{\mu_0 I}{R}$

   ›Reveal solutionSolution

   The magnetic field at the centre of a semicircular current-carrying arc is found using the Biot-Savart law; each element contributes perpendicular to the plane, and integration over the arc gives $\boxed{\frac{\mu_0 I}{4R}}$.

   

   The magnetic field at a point due to a current-carrying conductor is governed by the Biot-Savart law. For symmetric geometries like the circular arc shown in the figure, the calculation simplifies beautifully because every current element contributes in the same direction at the centre.

   When a current $I$ flows through a circular arc, each infinitesimal element $d\vec{l}$ produces a magnetic field at the centre that points perpendicular to the plane of the arc (determined by the right-hand rule). The key insight is that for a circular arc of radius $R$, the perpendicular distance from every element to the centre is exactly $R$, and the angle between $d\vec{l}$ and the position vector is always $90°$.

   For a circular arc subtending angle $\theta$ at its centre, the magnetic field is:

   $B = \frac{\mu_0 I \theta}{4\pi R}$

   Let me show you why this formula works and apply it to our semicircle.

   1. Apply the Biot-Savart law to a circular arc element For an element $d\vec{l}$ on the arc at distance $R$ from point O:

   $dB = \frac{\mu_0 I}{4\pi} \frac{|d\vec{l} \times \vec{r}|}{r^3} = \frac{\mu_0 I}{4\pi R^2} dl$

   Since $dl = R \, d\theta$ (arc length element), we have:

   $dB = \frac{\mu_0 I}{4\pi R} d\theta$

   2. Integrate over the semicircular arc A semicircle subtends an angle $\theta = \pi$ radians at its centre. All field contributions point in the same direction (into or out of the page), so: …

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3. CBSE 2025Set ANNUAL1 markMCQ
   Q.The magnetic field at the centre of a circular coil of radius 5 cm carrying a current of 1 A is 1.256 T. If the radius is made 10 cm, then the magnetic field at the centre of the loop carrying the same current will be

   (a) 2.512 T

   (b) 0.628 T

   (c) 5.024 T

   (d) 0.314 T

   ›Reveal solutionSolution

   For a circular coil, $B_{centre} = \mu_0 I/(2R)$, so $B \propto 1/R$ at fixed current; doubling $R$ halves $B$, giving $0.628$ T.

   Formula for field at the centre of a circular coil

   $B = \frac{\mu_0 I}{2R}$

   For a fixed current $I$, $B$ is inversely proportional to the radius $R$:

   $\frac{B_2}{B_1} = \frac{R_1}{R_2}$

   Substituting the given values

   $R_1 = 5$ cm, $B_1 = 1.256$ T, $R_2 = 10$ cm (current unchanged at 1 A):

   $B_2 = B_1 \times \frac{R_1}{R_2} = 1.256 \times \frac{5}{10} = 0.628\ \text{T}$

   …

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4. CBSE 2025Set ANNUAL1 markMCQ
   Q.Due to 10 amperes of current flowing in a circular coil of 10 cm radius, the magnetic field produced at its centre is 3.14 x 10^-3 weber/m^2. The number of turns in the coil will be

   (i) 5000

   (ii) 100

   (iii) 50

   (iv) 25

   ›Reveal solutionSolution

   N = B(2r)/(mu0 I) = 50 turns.

   The field at the centre of a coil of N turns is $B = \frac{\mu_0 N I}{2r}$. Solving for N: …

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5. CBSE 2024Set A1 markMCQ
   Q.The magnetic field produced at the centre of current carrying circular coil is (A) on the plane of coil (B) perpendicular to the plane of coil (C) at 45° to the plane of coil (D) at 180° to the plane of coil

   ›Reveal solutionSolution

   At the centre the magnetic field points along the axis — perpendicular to the plane of the coil.

   Using the Biot–Savart law, every current element of a circular coil contributes a field at the centre that points along the axis of the coil (perpendicular to its plane); by symmetry the in-plane components cancel. The magnitude at the centre is

   $B = \frac{\mu_0 N I}{2R}.$

   …

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6. CBSE 2024Set ANNUAL1 mark
   Q.A semi-circular arc of radius 20 cm carries a current of 10 A. What is the magnitude of the magnetic field at the centre of the arc? OR A galvanometer with a coil of resistance $12.0\ \Omega$ shows full-scale deflection for a current $2.5$ mA. How will you convert the galvanometer into an ammeter of range 0 to $7.5$ A?

   ›Reveal solutionSolution

   A semicircular arc contributes exactly half the field of a full circular loop of the same radius at the centre, since the field from a circular arc is proportional to the angle it subtends. The alternative converts a galvanometer to an ammeter using a low-resistance shunt in parallel.

   Field at the centre of a semicircular arc

   For a full circular loop of radius $R$ carrying current $I$, the field at the centre is $B_{loop} = \dfrac{\mu_0 I}{2R}$. A semicircular arc subtends half the angle ($\pi$ instead of $2\pi$) at the centre, so it contributes exactly half this field:

   $B = \frac{\mu_0 I}{4R}$

   Substituting $I = 10\,\text{A}$, $R = 20\,\text{cm} = 0.20\,\text{m}$:

   $B = \frac{(4\pi\times10^{-7})(10)}{4(0.20)} = \frac{4\pi\times10^{-6}}{0.8} = \frac{12.566\times10^{-6}}{0.8} = 1.57\times10^{-5}\,\text{T}$

   …

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7. CBSE 2024Set ANNUAL1 markMCQ
   Q.An infinitely long straight conductor is bent into the shape as shown in the figure. The magnetic field at the centre of the circular part is -

   (a) (μ₀/4π)·(2πI/R)

   (b) (μ₀/4π)·(2I/R)(π+1)

   (c) (μ₀/4π)·(2I/R)(π−1)

   (d) Zero

   ›Reveal solutionSolution

   The field at the centre is the sum of the field from the full circular loop and the field from the two straight semi-infinite wire segments (which together act like one infinite straight wire tangent to the circle).

   Field due to the circular loop (radius $R$, full turn):

   $B_{circle} = \dfrac{\mu_0 I}{2R}$

   Field due to the straight parts: The two straight segments are collinear and tangent to the circle at the junction point, so together they behave like a single infinite straight wire at perpendicular distance $R$ from the centre:

   $B_{straight} = \dfrac{\mu_0 I}{2\pi R}$

   …

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8. CBSE 2023Set ANNUAL1 markMCQ
   Q.A circular coil of radius 10 cm having 100 turns carries a current of 3.2 A. The magnetic field at the centre of the coil is

   (1) 2.01 x 10^-3 T

   (2) 5.64 x 10^-3 T

   (3) 2.64 x 10^-4 T

   (4) 5.64 x 10^-4 T

   ›Reveal solutionSolution

   Use the formula for the magnetic field at the centre of a circular current loop, scaled by the number of turns.

   $B=\dfrac{\mu_0 N I}{2R}=\dfrac{(4\pi\times10^{-7})(100)(3.2)}{2(0.1)}$

   …

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9. CBSE 2022Set ANNUAL1 mark
   Q.Draw a diagram of the magnetic field lines due to a current carrying circular loop.

   ›Reveal solutionSolution

   

   The magnetic field lines of a current-carrying circular loop form closed loops threading through the coil, straight along the central axis, and curving back around outside — the same pattern as the field of a bar magnet.

   Description of the field-line pattern for a circular current loop (current flowing, say, anticlockwise when viewed from the front):

   • Near the wire itself, the field lines are small concentric circles around the wire (as for any straight current element), by the right-hand rule.
   • At the centre of the loop and along its axis, the field lines are straight, directed along the axis (perpendicular to the plane of the loop), all pointing the same way (say, out of the page if current is anticlockwise). …

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10. CBSE 2022Set ANNUAL1 mark
    Q.Consider the circuit shown where APB and AQB are semicircles. What will be the magnetic field at the centre O of the circle?

    ›Reveal solutionSolution

    The two semicircular arcs carry equal currents in opposite senses about O, so their fields cancel: $B_O = 0$.

    The current $i$ entering at $A$ divides between the upper arc $APB$ and the lower arc $AQB$. Since the two semicircles have the same radius and hence the same length and resistance, the current divides equally: each carries $i/2$.

    The magnetic field at the centre of a semicircular arc of radius $R$ carrying current $I$ is

    $B = \frac{\mu_0 I}{4R}.$

    So each arc gives a field of magnitude

    $B_1 = B_2 = \frac{\mu_0 (i/2)}{4R} = \frac{\mu_0 i}{8R}.$

    …

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11. CBSE 2019Set ANNUAL1 markMCQ
    Q.Figure shows a circular coil with centre 'O'. If current flows through the coil, the magnetic field produced by it is maximum at :

    (a) Point O

    (b) Point P

    (c) Point K

    (d) Point N

    ›Reveal solutionSolution

    For a current-carrying circular coil, the magnetic field on its axis is strongest exactly at the centre and falls off as you move away along the axis.

    For a circular coil of radius $R$ carrying current $I$, the magnetic field at a point on its axis at distance $x$ from the centre is

    $B(x) = \frac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}}$

    …

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12. CBSE 2018Set ANNUAL1 markMCQ
    Q.The magnetic field at the centre of a circular coil of radius 5 cm carrying a current of 1 A is 1.256 T. If the radius is made 10 cm, then the magnetic field at the centre of the loop carrying same current is

    (a) 0.628 T

    (b) 2.512 T

    (c) 5.024 T

    (d) 0.314 T

    ›Reveal solutionSolution

    The field at the centre of a circular current loop scales as $1/r$ at fixed current, so doubling the radius halves the field.

    Step 1 — Field at centre of a circular loop

    $B = \frac{\mu_0 I}{2r} \implies B \propto \frac1r \ \text{(for fixed } I\text{)}$

    …

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