JEE Advanced 1987

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Animated Solution for Physics - Magnetic Effects of Current: A wire of length $L$ metre carrying a current $i$ ampere is bent in the form of circle. The magnitude of its magnetic moment is ...... in MKS units.

Q: What if the wire was wound into 2 turns instead of 1?

If wound into 2 turns, the circumference of each turn is $L/2$, making the radius $R' = L/(4\pi)$. The area of one turn is $A' = \pi R'^2 = L^2/(16\pi)$. The total magnetic moment for $N=2$ turns is $M = NiA' = 2i(L^2/(16\pi)) = iL^2/(8\pi)$, which is half of the single loop's moment!

Visualized Solution

$L \to Circle$

\text{Wire length} = L
\text{Current} = i

$M = i A$

M = iA
\text{where } A \text{ is the area of the loop}

$2 π R = L$

\text{Circumference} = 2\pi R = L
R = \frac{L}{2\pi}

$A = π R^{2}$

A = \pi R^2
A = \pi \left(\frac{L}{2\pi}\right)^2
A = \frac{L^2}{4\pi}

$M = \frac{i L ^{2}}{4 π}$

M = i \left(\frac{L^2}{4\pi}\right)
M = \frac{iL^2}{4\pi}

$N turns$

\text{What if the wire is wound into } n \text{ turns?}
M' = n \cdot i \cdot A'
M' = \frac{iL^2}{4\pi n}

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The Sigma Insight: Magnetic Moment of Current Loop

Solution Diagram

Bending the Wire

Imagine taking a straight wire of length

L

and bending it into a perfect circle. We are given that a steady current

i

flows through this circular loop. This simple geometric transformation is the key to unlocking the magnetic properties of the wire.

The Magnetic Moment Formula

We need to find the magnetic moment. The magnetic moment, denoted by

M

, for a planar current loop is simply the product of the current

i

and the area

A

it encloses. Mathematically, this is written as

M = i A

. This fundamental relationship tells us that to maximize the magnetic moment for a given current, we must maximize the enclosed area.

Finding the Area

To find the area, we first need the radius

R

. Since the entire wire of length

L

forms the boundary of the circle, the circumference

2 π R

must equal

L

. This gives us the radius

R = \frac{L}{2 π}

Now, let's calculate the area

A

. The area of a circle is

π R^{2}

. Substituting our expression for

R

, we get

A = π (\frac{L}{2 π})^{2}

. Simplifying this, the area becomes

A = \frac{L ^{2}}{4 π}

The Final Expression

Finally, let's substitute this area back into our magnetic moment formula. We get

M = i (\frac{L ^{2}}{4 π})

. And there we have it, the magnitude of the magnetic moment is

\frac{i L ^{2}}{4 π}

in MKS units. It is fascinating to see how a purely geometric constraint directly dictates the electromagnetic strength of the loop!

Common Traps & FAQs

Why is the magnetic moment just

i A

For a planar loop, the magnetic dipole moment is defined as the current multiplied by the vector area of the loop. It's a fundamental definition in electromagnetism that draws an analogy between current loops and magnetic dipoles.

Does the shape of the loop matter if the length is constant?

Yes! For a given perimeter (length of wire), a circle encloses the maximum possible area. Therefore, a circular loop will have the maximum magnetic moment compared to a square or triangular loop made from the same wire.

What does 'MKS units' mean here?

MKS stands for Meter-Kilogram-Second. It's the precursor to the SI unit system. In this context, it just means the answer should be in standard SI units (Ampere-meter squared), without any extra conversion factors.

What is the direction of this magnetic moment?

The direction is given by the right-hand grip rule. If you curl the fingers of your right hand in the direction of the current, your thumb points in the direction of the magnetic moment vector (perpendicular to the plane of the loop).

What if the wire was wound into 2 turns instead of 1?

If wound into 2 turns, the circumference of each turn is

L /2

, making the radius

R^{'} = L / (4 π)

. The area of one turn is

A^{'} = π R^{'2} = L^{2} / (16 π)

. The total magnetic moment for

N = 2

turns is

M = N i A^{'} = 2 i (L^{2} / (16 π)) = i L^{2} / (8 π)

, which is half of the single loop's moment!

Animated Solution for Physics - Magnetic Effects of Current: A wire of length $L$ metre carrying a current $i$ ampere is bent in the form of circle. The magnitude of its magnetic moment is ...... in MKS units.

Visualized Solution

$L \to Circle$

$M = i A$

$2 π R = L$

$A = π R^{2}$

$M = \frac{i L ^{2}}{4 π}$

$N turns$

The Sigma Insight: Magnetic Moment of Current Loop

Bending the Wire

The Magnetic Moment Formula

Finding the Area

The Final Expression

Similar Questions

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L→Circle

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M=iA

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2πR=L

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A=πR2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M=4πiL2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } N turns

The Sigma Insight: Magnetic Moment of Current Loop

Bending the Wire

The Magnetic Moment Formula

Finding the Area

The Final Expression

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A loop carrying current I lies in the x-y plane as shown in the figure. The unit vector k^ is coming out of the plane of the paper. The magnetic moment of the current loop is

$L \to Circle$

$M = i A$

$2 π R = L$

$A = π R^{2}$

$M = \frac{i L ^{2}}{4 π}$

$N turns$

A loop carrying current $I$ lies in the $x$ - $y$ plane as shown in the figure. The unit vector $\hat{k}$ is coming out of the plane of the paper. The magnetic moment of the current loop is