Question

Q] A point P lies on the axis of a flat coil carrying a current. The magnetic moment of the coil is μ .
What will be the magnetic field at point P? It is given that the distance of P from the centre of coil is
d, which is large compared to the radius of the coil.



​

Answer 1

Question:

A point P lies on the axis of a flat coil carrying a current. The magnetic moment of the coil is μ .

What will be the magnetic field at point P? It is given that the distance of P from the centre of coil is d, which is large compared to the radius of the coil.

Given:

Let O be the center of the coil.

Distance OP= d

Let radius be R

Angle between R and r= $\theta$

dB= magnetic field (it's not related to length d, it's just a notation)

Current=I

Magnetic moment = $\mu= NI \pi R^2$

Here N is number of turns in a coil. For a loop N=1

$\therefore \mu = I \pi R^2$

Solution:

We will use biot-savart law to derive the equation for the magnetic field created by the coil.

(See the attachment)

Since the current element is very small for a region dl, the magnetic field will also be small i.e dB and this dB will be perpendicular to the length OP.

$dB = \frac{ \mu_0} {4 \pi} \frac{I~dl~sin \phi} {r^2} \: \: \: \: \: \: [\phi ~is~the~angle~between~dl~and~ \vec{r} ~i.e~ sin 90°=1] \\\\ dB = \frac{ \mu_0} {4 \pi} \frac{I~dl} {r^2}- - - - [i]$

Now, break $d \vec{B}$ in its component i.e $dBcos \theta$ and $dB sin \theta$ as shown in the attachment.

So, the magnetic field we get along OP is $dBcos \theta$ for a small region of the loop dl.

But, We have to find the magnetic field for the whole circular loop.

So, we will integrate $dB~cos\theta$ for the whole loop.

$B_{net} = \int dB~cos \theta \\\\ \sf Putting~the~value~of~dB~from~[i], we~have: \\\\ B_{net} = \int \frac{ \mu_0} {4 \pi} \frac{I~dl} {r^2} cos \theta - - - - [ii]$

Find the value of $cos \theta$.

$Cos \theta = \frac{Adjacent}{Hypotenuse} \\\\ \sf See~the~attachment~\\\\ cos \theta = \frac{R} {r}$

Putting $cos \theta = \frac{R} {r}$ in [ii], we get the following:

$B_{net} = \int \frac{ \mu_0} {4 \pi} \frac{I~dl} {r^2} cos \theta \\\\ B_{net} = \int \frac{ \mu_0} {4 \pi} \frac{I~dl} {r^2} \bigg(\frac{R} {r} \bigg) \\\\ B_{net} = \frac{ \mu_0} {4 \pi} \frac{I}{r^2} \bigg(\frac{R} {r} \bigg) \int{dl}$

Integrating dl for the loop i.e from 0 to $2 \pi R$

$B_{net} = \frac{ \mu_0} {4 \pi} \frac{I}{r^2} \bigg(\frac{R} {r} \bigg) \int_{0}^{2 \pi R} {dl} \\\\ B_{net} = \frac{ \mu_0} {4 \pi} \frac{I}{r^2} \bigg(\frac{R} {r} \bigg) \bigg[dl \bigg]_{0}^{2 \pi R} \\\\ B_{net} = \frac{ \mu_0} {4 \pi} \frac{I}{r^2} \bigg(\frac{R} {r} \bigg) \bigg[2 \pi R - 0 \bigg] \\\\ B_{net} = \frac{ \mu_0} {4 \pi} \frac{I}{r^2} \bigg(\frac{R} {r} \bigg)2 \pi R \\\\B_{net} = \frac{ \mu_0} {4 \pi} \frac{2 \bf{I \pi R^2}}{r^3}$

But, We know that $I \pi R^2 = \mu$

$B_{net} = \frac{ \mu_0} {4 \pi} \frac{2 \bf{ \mu} }{r^3} \\\\ B_{net} = \frac{ \mu_0} {\cancel{4}~2\pi} \frac{\cancel{2}\bf{ \mu} }{r^3} \\\\ B_{net} = \frac{ \mu_0} {2\pi} \frac{\bf{ \mu} }{r^3}- - - - [iii]$

Now, let's simplify $r^3$ in [iii]

By, Pythagoras theorem, we have

$r^2= R^2+d^2 \\\\ r= \sqrt{R^2+d^2} \implies (R^2+d^2)^{\frac{1}{2}}\\\\ Put~ this~ value ~in ~[iii]$

$B_{net} = \frac{ \mu_0} {2\pi} \frac{\bf{ \mu} }{ \big((R^2+d^2)^{\frac{1}{2}} \big)^{3} } \\\\ B_{net} = \frac{ \mu_0} {2\pi} \frac{\bf{ \mu} }{ (R^2+d^2)^{\frac{3}{2}} }$

As per the question d>>R, so we will neglect R i.e R=0

$B_{net} = \frac{ \mu_0} {2\pi} \frac{\bf{ \mu} }{ (0+d^2)^{\frac{3}{2}} } \\\\ B_{net} = \frac{ \mu_0} {2\pi} \frac{\bf{ \mu} }{ (d^2)^{\frac{3}{2}} } \\\\ \boxed{B_{net} = \frac{ \mu_0} {2\pi} \frac{\bf{ \mu} }{d^3}}$

Answer 2

Answer:

Question:

A point P lies on the axis of a flat coil carrying a current. The magnetic moment of the coil is μ .

What will be the magnetic field at point P? It is given that the distance of P from the centre of coil is d, which is large compared to the radius of the coil.

Given:

Let O be the center of the coil.

Distance OP= d

Let radius be R

Angle between R and r= \thetaθ

dB= magnetic field (it's not related to length d, it's just a notation)

Current=I

Magnetic moment = \mu= NI \pi R^2μ=NIπR

2

Here N is number of turns in a coil. For a loop N=1

\therefore \mu = I \pi R^2∴μ=IπR

2

Solution:

We will use biot-savart law to derive the equation for the magnetic field created by the coil.

(See the attachment)

Since the current element is very small for a region dl, the magnetic field will also be small i.e dB and this dB will be perpendicular to the length OP.

\begin{gathered} dB = \frac{ \mu_0} {4 \pi} \frac{I~dl~sin \phi} {r^2} \: \: \: \: \: \: [\phi ~is~the~angle~between~dl~and~ \vec{r} ~i.e~ sin 90°=1] \\\\ dB = \frac{ \mu_0} {4 \pi} \frac{I~dl} {r^2}- - - - [i] \end{gathered}

dB=

4π

μ

0

r

2

I dl sinϕ

[ϕ is the angle between dl and

r

i.e sin90°=1]

dB=

4π

μ

0

r

2

I dl

−−−−[i]

Now, break d \vec{B}d

B

in its component i.e dBcos \thetadBcosθ and dB sin \thetadBsinθ as shown in the attachment.

So, the magnetic field we get along OP is dBcos \thetadBcosθ for a small region of the loop dl.

But, We have to find the magnetic field for the whole circular loop.

So, we will integrate dB~cos\thetadB cosθ for the whole loop.

\begin{gathered} B_{net} = \int dB~cos \theta \\\\ \sf Putting~the~value~of~dB~from~[i], we~have: \\\\ B_{net} = \int \frac{ \mu_0} {4 \pi} \frac{I~dl} {r^2} cos \theta - - - - [ii] \end{gathered}

B

net

=∫dB cosθ

Putting the value of dB from [i],we have:

B

net

=∫

4π

μ

0

r

2

I dl

cosθ−−−−[ii]

Find the value of cos \thetacosθ .

\begin{gathered} Cos \theta = \frac{Adjacent}{Hypotenuse} \\\\ \sf See~the~attachment~\\\\ cos \theta = \frac{R} {r} \end{gathered}

Cosθ=

Hypotenuse

Adjacent

See the attachment

cosθ=

r

R

Putting cos \theta = \frac{R} {r}cosθ=

r

R

in [ii], we get the following:

\begin{gathered}B_{net} = \int \frac{ \mu_0} {4 \pi} \frac{I~dl} {r^2} cos \theta \\\\ B_{net} = \int \frac{ \mu_0} {4 \pi} \frac{I~dl} {r^2} \bigg(\frac{R} {r} \bigg) \\\\ B_{net} = \frac{ \mu_0} {4 \pi} \frac{I}{r^2} \bigg(\frac{R} {r} \bigg) \int{dl}\end{gathered}

B

net

=∫

4π

μ

0

r

2

I dl

cosθ

B

net

=∫

4π

μ

0

r

2

I dl

(

r

R

)

B

net

=

4π

μ

0

r

2

I

(

r

R

)∫dl

Integrating dl for the loop i.e from 0 to 2 \pi R2πR

\begin{gathered} B_{net} = \frac{ \mu_0} {4 \pi} \frac{I}{r^2} \bigg(\frac{R} {r} \bigg) \int_{0}^{2 \pi R} {dl} \\\\ B_{net} = \frac{ \mu_0} {4 \pi} \frac{I}{r^2} \bigg(\frac{R} {r} \bigg) \bigg[dl \bigg]_{0}^{2 \pi R} \\\\ B_{net} = \frac{ \mu_0} {4 \pi} \frac{I}{r^2} \bigg(\frac{R} {r} \bigg) \bigg[2 \pi R - 0 \bigg] \\\\ B_{net} = \frac{ \mu_0} {4 \pi} \frac{I}{r^2} \bigg(\frac{R} {r} \bigg)2 \pi R \\\\B_{net} = \frac{ \mu_0} {4 \pi} \frac{2 \bf{I \pi R^2}}{r^3} \end{gathered}

B

net

=

4π

μ

0

r

2

I

(

r

R

)∫

0

2πR

dl

B

net

=

4π

μ

0

r

2

I

(

r

R

)[dl]

0

2πR

B

net

=

4π

μ

0

r

2

I

(

r

R

)[2πR−0]

B

net

=

4π

μ

0

r

2

I

(

r

R

)2πR

B

net

=

4π

μ

0

r

3

2IπR

2

But, We know that I \pi R^2 = \muIπR

2

=μ

\begin{gathered} B_{net} = \frac{ \mu_0} {4 \pi} \frac{2 \bf{ \mu} }{r^3} \\\\ B_{net} = \frac{ \mu_0} {\cancel{4}~2\pi} \frac{\cancel{2}\bf{ \mu} }{r^3} \\\\ B_{net} = \frac{ \mu_0} {2\pi} \frac{\bf{ \mu} }{r^3}- - - - [iii] \end{gathered}

B

net

=

4π

μ

0

r

3

2μ

B

net

=

4

2π

μ

0

r

3

2

μ

B

net

=

2π

μ

0

r

3

μ

−−−−[iii]

Now, let's simplify r^3r

3

in [iii]

By, Pythagoras theorem, we have

\begin{gathered} r^2= R^2+d^2 \\\\ r= \sqrt{R^2+d^2} \implies (R^2+d^2)^{\frac{1}{2}}\\\\ Put~ this~ value ~in ~[iii] \end{gathered}

r

2

=R

2

+d

2

r=

R

2

+d

2

⟹(R

2

+d

2

)

2

1

Put this value in [iii]

\begin{gathered}B_{net} = \frac{ \mu_0} {2\pi} \frac{\bf{ \mu} }{ \big((R^2+d^2)^{\frac{1}{2}} \big)^{3} } \\\\ B_{net} = \frac{ \mu_0} {2\pi} \frac{\bf{ \mu} }{ (R^2+d^2)^{\frac{3}{2}} } \end{gathered}

B

net

=

2π

μ

0

((R

2

+d

2

)

2

1

)

3

μ

B

net

=

2π

μ

0

(R

2

+d

2

)

2

3

μ

As per the question d>>R, so we will neglect R i.e R=0

\begin{gathered} B_{net} = \frac{ \mu_0} {2\pi} \frac{\bf{ \mu} }{ (0+d^2)^{\frac{3}{2}} } \\\\ B_{net} = \frac{ \mu_0} {2\pi} \frac{\bf{ \mu} }{ (d^2)^{\frac{3}{2}} } \\\\ \boxed{B_{net} = \frac{ \mu_0} {2\pi} \frac{\bf{ \mu} }{d^3}} \end{gathered}

B

net

=

2π

μ

0

(0+d

2

)

2

3

μ

B

net

=

2π

μ

0

(d

2

)

2

3

μ

B

net

=

2π

μ

0

d

3

μ