Learn Biot Savart's law derivation for a point charge and current element. Practice to calculate the magnetic field due to current carrying ring at a point placed on its axis and section of ring at the center.

- Magnetic field is created due to motion of charge.
- Biot Savart's law gives the value of magnetic field due to moving point charge.
- To understand Biot Savart's law, consider a positive point charge (q) moving with a velocity \(\vec v\) as shown in figure.

- Magnetic field vector at a point P at a position vector \(\vec r\) from the point charge will be

\(\vec B = \dfrac {\mu_0} {4\pi} q \dfrac {(\vec v × \hat r)} {r^2}\) \(Tesla\)

\(\vec B = \dfrac {\mu_0} {4\pi} q \dfrac {|\vec v| |\hat r|} {r^2} sin \theta \)

\(\vec B \) = \( \dfrac {\mu_0} {4\pi} q \dfrac {vsin\theta} {r^2} \) \(Tesla \) [\(\because\) \(|\hat r|=1\)]

A \(2×10^9 \,(\hat{k})\,T\)

B \(4× 10^{-7}\, (\hat{j})\,T\)

C \(1.15×10^{-17} \,(-\hat{k})\,T\)

D \(0 \;T\)

- Consider a current carrying conductor as shown in figure.

- Consider a small element AB of current carrying wire whose length is \(d\vec s\) and the position vector of point (P) from the element is \(\vec r\).
- Magnetic field at point P due to element AB can be calculated by Biot Savart's law.

- According to Biot Savart's law if the magnetic field vector at point P due to current carrying element is \(d\vec B\), then

\(\rightarrow \,\,d\vec B\) is perpendicular to both \(d\vec s\) and \(\hat r\)(unit vector along \(\vec r\)).

\(\rightarrow\) \(d\vec B\) is proportional to current and to the magnitude of the length element \(d\vec s\).

\(\ d\vec B \propto I\,\, |d\vec s|\)

\(\rightarrow\) The magnitude of \(d\vec B\) is proportional to \(sin \theta\), where \(\,\theta\) is the angle between the vectors \(d\vec s\) and \(\hat r\).

\(|d\vec B| \propto sin\theta\)

\(\rightarrow\) \(d\vec B\) is inversely proportional to \(r^2\) , where r is the distance between \(d\vec s\) and P.

\(|d\vec B|\propto \dfrac{1}{r^2}\)

- Thus the magnetic field at point (P) is given as

\(d\vec B=\dfrac {\mu_0}{4\pi} I \dfrac {(d\vec s × \hat r)} {r^2}\) or \(d\vec B=\dfrac {\mu_0}{4\pi} I \dfrac {ds\, sin\theta} {r^2}\)[direction of magnetic field is given by \((d\vec s × \hat r)\)]

This mathematical expression is called Biot - Savart's law.

- Here, \(\mu_0\) is a physical constant which is called permitivity of free space.

\(\mu_0 = 4\pi×10^{-7}\) \( T m/A \)

A \(1.8×10^{-9} (\hat k)\,T\)

B \(2×10^{-9} (-\hat k)\,T \)

C \(3×10^{-9} (-\hat k) \,T\)

D \(4×10^{-8} (\hat k)\,T\)

- Consider a finite conducting rod of length L carrying a current \(I\) is placed along x-axis as shown in figure.

- To calculate the magnetic field at point P situated at a distance 'd' on y-axis, consider a small element of length \(d\vec \ell\) located at a distance r from P.
- The direction of magnetic field at point P due to the small current carrying element is outward means along \(\hat k\).[using right hand thumb rule]

Magnetic field \(d\vec B\) due to small element at point P is

\(d\vec B = \dfrac {\mu_0} {4\pi} I \dfrac {d\vec \ell × \hat r} {r^2}\)

\(d\vec B = \dfrac {\mu_0} {4\pi} \dfrac {I}{r^2} d\ell sin\)\((\dfrac {\pi} {2} - {\theta)} \hat k\)

\(d\vec B = \dfrac {\mu_0} {4\pi} \dfrac {Id\ell cos\theta} {r^2} {\hat k} \)....(1)

From figure,

\(tan\theta =\dfrac {-\ell}{d}\)

or, \(\ell = -d\,\tan\theta\)

Differentiating both sides

\(d\ell = -dsec^2 \theta d \theta\)

Also, from figure

\(cos\theta=\dfrac {d}{r}\)

or, \(r\) = \(\dfrac {d}{cos\theta}\)

Put the value of \(d\ell\) and \(r\) in equation (1)

\(d\vec B = \dfrac {\mu_0} {4\pi} I \)\(\dfrac {(-dsec^2\theta d\theta × cos\theta × cos^2\theta)}{d^2}\)

\(d \vec B = - \dfrac {\mu_0} {4\pi d} I cos\theta d\theta \)

Integrating both sides

\(\int d\vec B = -\dfrac {\mu_0I} {4\pi d} \int^{-\theta_2} _{\theta_1} cos\theta d\theta \)

or \( \vec B = -\dfrac {\mu_0I} {4\pi d} \)\([sin\theta]^{-\theta_2}_{\theta_1}\)

\( \vec B = -\dfrac {\mu_0I} {4\pi d} \)\([-sin\theta_2-sin\theta_1]\)

\( \vec B = \dfrac {\mu_0I} {4\pi d} \)\([sin\theta_1+sin\theta_2]\)

A \(10^{-8} \,T \) outward the plane

B \(10^{-7}\, T \) outward the plane

C \(13×10^{-8} T \) outward the plane

D \(10^{-10}\, T \) inward the plane

- Consider a section of ring of radius \(R\) having current \(I\) as shown in figure.

- Consider a small element of length \(d\ell\) making an angle \(d\theta\).

- The magnetic field at point \(O\) due to small element \(d\ell\) is given as

\( dB = \dfrac {\mu_0I d\ell}{4\pi R^2}\)

- From figure

\(d\theta = \dfrac{d\ell}{R}\) \(\Bigg[Angle =\dfrac {Arc}{Radius}\Bigg]\)

\(d\ell = Rd\theta\)

so, magnetic field will be

\( dB = \dfrac {\mu_0I Rd\theta}{4\pi R^2}\)

\( \int dB = \dfrac {\mu_0I }{4\pi R}\int^{\theta}_0 d\theta\)

\( \vec B = \dfrac {\mu_0I \theta}{4\pi R}\) [ \(\theta\,\)must be in radian]

\( \vec B = \dfrac {\mu_0I \theta}{4\pi R}\)(Direction inside the plane)

A \(3\pi × 10^{-4} \hat k \,T\)

B \(8\pi × 10^{3} \hat k \,T\)

C \(5\pi × 10^{-6} \hat k \,T\)

D \(0\)

- Consider a ring of radius R placed in y-z plane as shown in figure.
- To calculate magnetic field at point P at x - distance from the center on its axis, consider the elements as shown in figure.

Magnetic field due to small element, \(dB = \dfrac {\mu_0I}{4\pi} \dfrac{d\ell}{r^2}\)

- From figure, it is clear that vertical components of magnetic field get cancelled and horizontal components get added.
- Hence, the net magnetic field at P will be along positive x - direction.

\(dB_x = \dfrac {\mu_0I} {4\pi} \dfrac {d\ell}{R^2+x^2}\) \(cos\theta\) [\(\because\,\)r = \(\sqrt {R^2 + x^2}\)]

\(dB_x = \dfrac {\mu_0I} {4\pi} \dfrac {d\ell}{R^2+x^2} \dfrac {R}{\sqrt{R^2+x^2} }\) \(\Bigg[\because\, cos\theta=\dfrac {R} {\sqrt{R^2+x^2}}\Bigg]\)

\(dB_x = \dfrac {\mu_0I} {4\pi} \dfrac {d\ell R}{(R^2+x^2)^{3/2}} \)

Integrating both sides

\(\int dB_x = \dfrac {\mu_0IR} {4\pi{(R^2+x^2)^{3/2}}} \int d\ell \)

\(\int dB_x = \dfrac {\mu_0IR} {4\pi{(R^2+x^2)^{3/2}}} \int^{2\pi}_{0} Rd\theta \) \([d\ell = Rd\theta]\)

\(B_x = \dfrac {\mu_0IR^2} {4\pi{(R^2+x^2)^{3/2}}} [2\pi]\)

or ,\(B_x = \dfrac {\mu_0IR^2} {2{(R^2+x^2)^{3/2}}}\)

in terms of \(\theta\),

\(B_x = \dfrac {\mu_0IR\, sin\theta} {2{(R^2+x^2)^{}}}\)

A \(2.2 \pi \, T\)

B \(2.54\,\pi × 10^{-3} \, T\)

C \(2.88\,\pi × 10^{-6} \, T\)

D \(2.3 \,\pi× 10^{9} \, T\)

A \(4\sqrt 2 × 10^{-5}\, T\)

B \(3\sqrt 2 × 10^{-5}\, T\)

C \(8\sqrt 2 × 10^{-5}\, T\)

D \(2\sqrt 2 × 10^{-5}\, T\)

A \(1\,\)\(T\)

B \(2\pi \,\,T\)

C \(0\,T\)

D \(2\pi × 10^{-5}\, T\)

A \(4×10^{-5}\, T (outwards)\)

B \(6×10^{-5} \,T (outwards)\)

C \(2×10^{-5} \,T (inwards)\)

D \(2×10^{-5}\, T (outwards)\)