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PhysicsElectromagnetism / Magnetic Effects

A circular loop of radius $r$ carries current $I_0$. The magnetic field at the centre of the loop due to a $60°$ arc is:

R
Solution written and verified by Roshan, science educator with 5 years of experience teaching NEET and JEE aspirants. Last reviewed September 2026.
Options
1
$\dfrac{\mu_0 I_0}{12r}$
2
$\dfrac{\mu_0 I_0}{6r}$
3
$\dfrac{\mu_0 I_0}{4r}$
4
$\dfrac{\mu_0 I_0}{3r}$
Correct Answer
$\dfrac{\mu_0 I_0}{12r}$
Solution
1

Full circular loop: $B = \mu_0 I_0/2r$

$60°$ arc = $60/360 = 1/6$ of full loop

2

$B_{60°} = \frac{1}{6} \times \frac{\mu_0 I_0}{2r} = \frac{\mu_0 I_0}{12r}$

Answer: $\dfrac{\mu_0 I_0}{12r}$

Full loop field = $\mu_0 I/2r$; fraction = arc angle/360°
60° arc: B = (1/6)×(μ₀I/2r) = μ₀I/12r
Theory: Electromagnetism / Magnetic Effects
1. Magnetic Effects of Current

Biot-Savart law: $dB = \mu_0 I dl\sin\phi / 4\pi r^2$. Ampere\'s law: $\oint B\cdot dl = \mu_0 I_{enc}$. Direction: right-hand rule. Current-carrying conductor in field: $\vec{F} = I(\vec{L}\times\vec{B})$ (force on wire). Force between parallel wires: $F/L = \mu_0 I_1 I_2/2\pi d$ (attractive if same direction, repulsive if opposite). This defines the SI unit of Ampere.

2. Moving Charge in Magnetic Field

Lorentz force: $\vec{F} = q\vec{v}\times\vec{B} = qvB\sin\theta$. Circular motion: $qvB = mv^2/r$ → $r = mv/qB$. Cyclotron frequency: $f = qB/2\pi m$ (independent of speed). Velocity selector (crossed E and B): $E = vB$ → $v = E/B$. Mass spectrometer: $r = mv/qB$ → different masses separate.

3. Magnetic Moment and Torque

Magnetic dipole moment: $\vec{m} = NIA\hat{n}$ (for coil of N turns, area A). Torque: $\vec{\tau} = \vec{m}\times\vec{B} = mB\sin\theta$. Potential energy: $U = -\vec{m}\cdot\vec{B} = -mB\cos\theta$. Moving coil galvanometer: $\tau_{magnetic} = nBIA$, $\tau_{spring} = k\theta$ → $\theta = nBIA/k$ (deflection proportional to current).

4. Electromagnetic Induction

Faraday\'s law: $\varepsilon = -d\Phi_B/dt$ where $\Phi_B = \int B\cdot dA$ (magnetic flux). Lenz\'s law: induced current opposes change in flux (ensures energy conservation). Motional EMF: $\varepsilon = Blv$ (conductor of length $l$ moving at $v$ perpendicular to $B$). Self-inductance: $\varepsilon = -L\, dI/dt$; $L = N\Phi/I$. Solenoid: $L = \mu_0 N^2 A/l = \mu_0 n^2 Al$. Energy stored: $U = \frac{1}{2}LI^2$.

5. Force on a Current-Carrying Conductor

A wire of length L carrying current I in a field B feels $F = BIL\sin\theta$, where $\theta$ is the angle between the wire and the field. The direction follows Fleming's left-hand rule. Between two parallel wires the force per unit length is $\frac{\mu_0 I_1 I_2}{2\pi d}$ — attractive when the currents flow the same way and repulsive when they oppose, which is the opposite of what students expect from charges, and is the definition on which the ampere itself was based.

6. The Moving Coil Galvanometer

A coil of N turns and area A carrying current I in a radial field experiences a torque $NBIA$, balanced by the restoring torque $k\phi$ of the suspension, giving $\phi = \frac{NBA}{k}I$ — a deflection directly proportional to current. Sensitivity rises with N, B and A, and falls with k. A galvanometer becomes an ammeter by adding a small shunt resistance in parallel, and a voltmeter by adding a large resistance in series.

7. Magnetic Materials

Diamagnetic substances are weakly repelled and have no unpaired electrons, with relative permeability just below 1. Paramagnetic substances have unpaired electrons and are weakly attracted. Ferromagnetic substances contain domains that align strongly with the field, giving permeability in the thousands, and they retain magnetisation after the field is removed — the hysteresis that makes permanent magnets possible. Heating a ferromagnet past its Curie temperature destroys the domain alignment and it becomes paramagnetic.

Where students lose the mark

Confusing the left- and right-hand rules. Fleming's left hand gives the force on a current in a field; the right hand gives the induced current when a conductor moves.

Forgetting $\sin\theta$. A wire parallel to the field feels no force at all, however large the current.

Frequently Asked Questions
1. What is the field at centre of a full circular loop? ⌄
$B = \mu_0 I / 2R$ at centre of circular loop of radius $R$ carrying current $I$. Derived from Biot-Savart law: each element $dl$ contributes $dB = \mu_0 I dl / 4\pi r^2$; integrating over full circle.
2. How do you find field due to an arc? ⌄
Field due to arc = (angle of arc / 2$\pi$) × field of full circle. For arc of angle $\theta$ (in radians): $B_{arc} = (\theta/2\pi) \times \mu_0 I/2R = \mu_0 I\theta/4\pi R$. For $60^\circ = \pi/3$ radians: $B = \mu_0 I (\pi/3)/4\pi R = \mu_0 I/12R$.
3. What is Biot-Savart law? ⌄
$d\vec{B} = \frac{\mu_0}{4\pi}\frac{I\,d\vec{l}\times\hat{r}}{r^2}$. The magnetic field due to a current element $I\,dl$ at position $\vec{r}$ away from it. $\mu_0 = 4\pi\times10^{-7}$ T m A$^{-1}$ (permeability of free space).
4. What are other magnetic field results? ⌄
Infinite straight wire: $B = \mu_0 I/2\pi r$. Finite wire: use Biot-Savart integration. Solenoid (infinite): $B = \mu_0 nI$ inside, 0 outside. Toroid: $B = \mu_0 NI/2\pi r$ inside. On axis of circular loop at distance $x$: $B = \mu_0 IR^2/2(R^2+x^2)^{3/2}$.
5. What is Ampere's law? ⌄
$\oint \vec{B}\cdot d\vec{l} = \mu_0 I_{enc}$. Useful for high symmetry. For infinite straight wire: $B(2\pi r) = \mu_0 I$ → $B = \mu_0 I/2\pi r$. For toroid: $B(2\pi r) = \mu_0 NI$ → $B = \mu_0 NI/2\pi r$. For solenoid of $n$ turns/m: $B \cdot L = \mu_0 nLI$ → $B = \mu_0 nI$.
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