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Moving Charges and Magnetism
Magnetic Dipole Moment of Current Loop
5 questions from 5 papers, each with its answer.
- questions
- 5
- papers
- 5
- years asked, 2017–2026
- 3 of 10
- of Moving Charges and Magnetism
- 5.62%
Appears in
Concepts tested
- Magnetic Dipole Moment
- Magnetic moment of a circulating charge
- Magnetic moment of a coil
- Magnetic moment of a current-carrying circular loop
- magnetic-dipole-work-rotation
All Questions
2025 1 question
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A $2\text{ amp}$ current is flowing through two different small circular copper coils having radii ratio $1:2$. The ratio of their respective magnetic moments will be:
Magnetic moment is $M = NIA = I(\pi r^2) \propto r^2$. Thus $M_1/M_2 = (r_1/r_2)^2 = (1/2)^2 = 1/4$.
The magnetic dipole moment of a current loop is $M = N I A = N I (\pi r^2)$. Since the current $I$ and number of turns $N$ are identical, the magnetic moment is directly proportional to the square of the coil radius: $M \propto r^2$. Therefore, $\frac{M_1}{M_2} = \left(\frac{r_1}{r_2}\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$.
2020 Phase-2 1 question
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A wire of length L metre carrying a current of I ampere is bent in the form of circle. Its magnetic moment isCircumference $L=2\pi r\Rightarrow r=\dfrac{L}{2\pi}$. Magnetic moment $M=I\times\pi r^2=I\pi\left(\dfrac{L}{2\pi}\right)^2=\dfrac{IL^2}{4\pi}$.
2017 1 question
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A 250-Turn rectangular coil of length 2.1 cm and width 1.25 cm carries a current of 85 $\mu A$ and subjected to a magnetic field of strength 0.85 T. Work done for rotating the coil by $180^\circ$ against the torque is
Rotating from $0^\circ$ to $180^\circ$ requires $W = 2NIAB$.
$W = MB(\cos\theta_1 - \cos\theta_2)$ When it is rotated by $180^\circ$, $W = 2MB$ $W = 2(NIA)B$ $= 2\times250\times85\times10^{-6}\times[1.25\times2.1\times10^{-4}]\times85\times10^{-2}$ $= 9.1\ \mu J$
2007 AIPMT-PRE 1 question
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A charged particle (charge $q$) is moving in a circle of radius $R$ with uniform speed $v$. The associated magnetic moment $\mu$ is given byA charge moving in a circle is equivalent to a current loop.
Time period of revolution: $T = \dfrac{2\pi R}{v}$
Equivalent current: $I = \dfrac{q}{T} = \dfrac{qv}{2\pi R}$
Magnetic moment: $\mu = IA$, with area $A = \pi R^2$
$\mu = \dfrac{qv}{2\pi R}(\pi R^2)$
$\mu = \dfrac{qvR}{2}$
2001 AIPMT 1 question
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If number of turns, area and current through a coil are given by n, A and i respectively then its magnetic moment will be : -Magnetic moment of a single current loop: $m = iA$.
For a coil of n turns, each turn contributes the same moment, so $M = niA$.
Its direction is along the axis of the coil, given by the right hand rule.