Informative line

Biot Savarts Law

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.

Biot Savart's Law for a Point Charge

  • 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 

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

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

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

Illustration Questions

A proton situated at (3,4) is moving along +y direction with a velocity of \(\vec v = 3 × 10^6\) m/s . Calculate the magnetic field vector at the origin due to proton.  [ \(\dfrac {\mu_0} {4\pi} = 10^{-7}\) \(Tm/A\) and charge on proton = \( 1\cdot6 ×10^{-19} C\)]

A \(2×10^9 \,(\hat{k})\,T\)

B \(4× 10^{-7}\, (\hat{j})\,T\)

C \(1.15×10^{-17} \,(-\hat{k})\,T\)

D \(0 \;T\)

×

From Biot Savart's law,

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

Where, \(q\) = Charge of proton

\(\vec v\)= velocity of proton= \(3×10^6 \hat j \,\,m/s\)

\(\hat r\)= position vector

image

Position vector \(|\vec r|\)

\(\vec r = (x_2-x_1) \hat i + (y_2-y_1) \hat j \)

\(\vec r = +3 \hat i + 4 \hat j \)

\(|\vec r| = \sqrt {(3)^2+(4)^2} =5\,cm\)

\(\hat r = \dfrac {\vec r} {|\vec r|}=\dfrac {3\hat i + 4 \hat j} {5}\)

image

Magnetic field at origin,

\(\vec B_{(0,0)} = \dfrac {\mu_0} {4\pi} × 1\cdot6 ×10^{-19}\)× \(\Bigg[\dfrac {3×10^{6} \hat j × \dfrac{(3\hat i +4\hat j)} {5}} {(5×10^{-2})^2}\Bigg]\)

\(\vec B_{(0,0)} \)\(10^{-7} × 1.6 × 10^{-19} \) × \(\dfrac {9×10^6(-\hat k)}{125×10^{-4}} = 1.15× 10^{-17}(-\hat k)\, T\)

image

A proton situated at (3,4) is moving along +y direction with a velocity of \(\vec v = 3 × 10^6\) m/s . Calculate the magnetic field vector at the origin due to proton.  [ \(\dfrac {\mu_0} {4\pi} = 10^{-7}\) \(Tm/A\) and charge on proton = \( 1\cdot6 ×10^{-19} C\)]

image
A

\(2×10^9 \,(\hat{k})\,T\)

.

B

\(4× 10^{-7}\, (\hat{j})\,T\)

C

\(1.15×10^{-17} \,(-\hat{k})\,T\)

D

\(0 \;T\)

Option C is Correct

Biot Savart's law for Current Element

  • 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.

         \(\mu_0 = 4\pi×10^{-7}\) \( T m/A \)

Illustration Questions

The current \(I = 4\,A\) is flowing along a positive y direction in a rod placed along y-axis. Find the magnetic field due to an element of rod of length \(d\ell\) = 5 cm at a point (in x-y plane) placed at a distance of  r = 2m and making an angle \(\theta\)=37° with element of rod. [\(\because\) \( sin 37°=\)\(\dfrac {3}{5}\)]

A \(1.8×10^{-9} (\hat k)\,T\)

B \(2×10^{-9} (-\hat k)\,T \)

C \(3×10^{-9} (-\hat k) \,T\)

D \(4×10^{-8} (\hat k)\,T\)

×

Magnetic field due to small element of length \(d\vec l \) at a point placed at distance \(\vec r\) is given as

\(d\vec B=\dfrac {\mu_0}{4\pi} I \dfrac {d\vec l × \vec r} {r^3}\)

 

image

Since, the distance r is much larger than the length of element of rod.

So, 

\(d\vec B=\dfrac {\mu_0}{4\pi} I \dfrac {d l\, sin\theta} {r^2}\)

 

image

Given: \(I = 4\,A\,,\)   \(d\ell \)\(=5\,cm\,,\,\,\,\,\, r = 2\, m\,,\)  \(\theta=37°\)

\(d\vec B=\dfrac {10^{-7}×4×5×{10^{-2}× sin37°}} {(2^2)}\)

\(d\vec B = 3×10^{-9}\,\)\(T\)

image

The direction of magnetic field is given by the cross - product of \(d\vec\ell×\vec r\).

So, by using right hand thumb rule, direction of magnetic field will be along \(-\hat k \).

image

The current \(I = 4\,A\) is flowing along a positive y direction in a rod placed along y-axis. Find the magnetic field due to an element of rod of length \(d\ell\) = 5 cm at a point (in x-y plane) placed at a distance of  r = 2m and making an angle \(\theta\)=37° with element of rod. [\(\because\) \( sin 37°=\)\(\dfrac {3}{5}\)]

image
A

\(1.8×10^{-9} (\hat k)\,T\)

.

B

\(2×10^{-9} (-\hat k)\,T \)

C

\(3×10^{-9} (-\hat k) \,T\)

D

\(4×10^{-8} (\hat k)\,T\)

Option C is Correct

Magnetic Field at a Point due to Finite Rod

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

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

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

 

Illustration Questions

A long straight rod of length \(L = 4\,m\)  carrying current \(I=2\,A \) is placed at a distance of \( d = 2\, m\) from point P making an angle with both ends of rod ,such that, \(\theta_1=30°\) and \(\theta_2=53°\). Calculate the magnetic field at point P and its direction. [ \(sin\) \(53\)° = \(\dfrac {4} {5}\)]

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

×

Magnetic field at point P due to finite rod of length L is given as 

\(\vec B = \dfrac {\mu_0 I} {4\pi d} [sin\theta_1+sin\theta_2]\)

image

Given : \(\text I\)\(= 2\,A\,,\,\,\,d = 2\,m\,, \)  \(\theta_1=30°\,,\)  \(\theta_2=53°\)

\(\vec B = 10^{-7} ×\dfrac {2}{2}\)[\(sin \,30° + sin \,53°\)]

\(\vec B = 10^{-7} [\dfrac {1}{2}+\dfrac {4}{5}]\)

\(\vec B = 13×10^{-8}\, T\)

 

image

Using right hand thumb rule, direction of magnetic field is outward.

image

A long straight rod of length \(L = 4\,m\)  carrying current \(I=2\,A \) is placed at a distance of \( d = 2\, m\) from point P making an angle with both ends of rod ,such that, \(\theta_1=30°\) and \(\theta_2=53°\). Calculate the magnetic field at point P and its direction. [ \(sin\) \(53\)° = \(\dfrac {4} {5}\)]

image
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

Option C is Correct

Magnetic Field due to Section of Ring at the Center

  • 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)

 

 

 

Illustration Questions

A quarter part of a current carrying ring of radius R = 2 cm is placed in x - y plane as shown in figure in which \(I=2\,A\) of current is flowing. Calculate the magnetic field vector at the center of the circle.

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 \(0\)

×

Magnetic field due to section of ring of radius \(R\) at its center is given as 

\( \vec B = \dfrac {\mu_0I \theta}{4\pi R} \hat k\)    [Direction of magnetic field is in \(\hat k\) using right hand rule]

where, \(\theta = \dfrac {\pi} {2}\)

 

image

Given : \( R = 2 \,cm\,,\)   \(\theta = \dfrac {\pi} {2}\,,\)    \(I\)\(= 2\,A \)

\( \vec B = \dfrac {10^{-7} × 2 × \pi} {2×10^{-2} × 2} \hat k\)

\( \vec B = \dfrac {\pi} {2}× 10^{-5}\hat k\)

\( \vec B = {5\pi} × 10^{-6}\hat k\,\, T\)

 

image

A quarter part of a current carrying ring of radius R = 2 cm is placed in x - y plane as shown in figure in which \(I=2\,A\) of current is flowing. Calculate the magnetic field vector at the center of the circle.

image
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

\(0\)

Option C is Correct

Magnetic Field due to Current Carrying Ring at a Point Placed on its Axis

  • 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 \(\theta\),

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

Illustration Questions

Calculate the magnetic field due to ring of radius R = 3 cm carrying a current \(I\)= 2 A on point P at a distance x = 4 cm from the center along the axis.

A \(2.2 \pi \, T\)

B \(2.54\,\pi × 10^{-3} \, T\)

C \(2.88\,\pi × 10^{-6} \, T\)

D \(2.3 \,\pi× 10^{9} \, T\)

×

Magnetic field due to a ring at a point on its axis is given as 

B = \(\dfrac {\mu_0 IR^2}{2(R^2+x^2)^{3/2}}\)

image

Given: \(R = 3 \,cm\) ,   \(I\)\( = 2\,A \,\,\,\,,\,\,x = 4 \,cm \)

 \(B=\dfrac {4 \pi ×10^{-7} × 2 × (3× 10^{-2})^2} {2[(3^2+4^2) × 10^{-4}]^{3/2}} \)

\(B=2.88\,\pi × 10^{-6} \ T\)

image

Calculate the magnetic field due to ring of radius R = 3 cm carrying a current \(I\)= 2 A on point P at a distance x = 4 cm from the center along the axis.

image
A

\(2.2 \pi \, T\)

.

B

\(2.54\,\pi × 10^{-3} \, T\)

C

\(2.88\,\pi × 10^{-6} \, T\)

D

\(2.3 \,\pi× 10^{9} \, T\)

Option C is Correct

Illustration Questions

A current \(I\)\( = 2\,A \) is flowing in a square loop of side \(a = 2 \,cm\,.\) Calculate the magnetic field vector at the center of loop .

A \(4\sqrt 2 × 10^{-5}\, T\)

B \(3\sqrt 2 × 10^{-5}\, T\)

C \(8\sqrt 2 × 10^{-5}\, T\)

D \(2\sqrt 2 × 10^{-5}\, T\)

×

Magnetic field at the center of the loop will be vector sum of magnetic field due to individual sides of the square loop.

image

Magnetic field at O due to wire AB

\(\vec B_{AB} =\)\(\dfrac{\mu_0I}{4\pi\,d} (sin\alpha+sin\beta)\) , where \(d=\dfrac{a}{2}\)

\(\vec B_{AB} =\dfrac{10^{-7}× {2 [sin 45° + sin45°]}}{(1×10^{-2})} \)

\(\vec B_{AB} \)\(= 2\sqrt 2 × 10^{-5} T\) [outwards the paper]

 

image image

Similarly, magnetic field at \(O\) due to wire BC,

\(\vec B_{BC} \) = \(2\sqrt 2 × 10^{-5}\, T\)  [outwards the paper]

Magnetic field due to wire CD ,

\(\vec B_{CD} \) = \(2\sqrt 2 × 10^{-5}\, T\)  [outwards the paper]

Magnetic field due to wire DA ,

\(\vec B_{DA} \)\(2\sqrt 2 × 10^{-5}\, T\)  [outwards the paper]

image

Hence, net magnetic field at O due to loop ,

\(\vec B =\)\(\vec B_{AB} +\vec B_{BC} +\vec B_{CD} +\vec B_{DA} \)

\(\vec B \) = \(8\sqrt 2 × 10^{-5}\, T\)

image

A current \(I\)\( = 2\,A \) is flowing in a square loop of side \(a = 2 \,cm\,.\) Calculate the magnetic field vector at the center of loop .

image
A

\(4\sqrt 2 × 10^{-5}\, T\)

.

B

\(3\sqrt 2 × 10^{-5}\, T\)

C

\(8\sqrt 2 × 10^{-5}\, T\)

D

\(2\sqrt 2 × 10^{-5}\, T\)

Option C is Correct

Illustration Questions

Calculate the magnetic field at point O due to current carrying element with current \(I\) = 2A as shown in figure.[Given r = 1 cm] [Assume the plane of the paper as x-y plane]

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

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

C \(0\,T\)

D \(2\pi × 10^{-5}\, T\)

×

Since, the figure consists of three parts AB, BC and CD.

So, the total magnetic field at O can be calculated by taking vector sum of magnetic field due to individual elements.

image

Magnetic field at O due to AB ,

Since for every small elements of AB, \(d\vec \ell\)  and \(\vec r\) are parallel.

So, \(d\vec \ell × \vec r = 0\)

Hence, \(\vec B_{AB} = 0\)

image image

Magnetic field at O due to CD ,

Since, for every small element of CD \(d\vec \ell \) and \(\vec r \) are parallel .

So, \(\vec B_{CD} = 0\)

image image

Magnetic field at O due to BC 

\(\vec B _{BC}= \dfrac {\mu_0}{4\pi r} I\theta \)

where \(I\) = current,

\(\theta\)=angle between B and C,

\(r \)= radius / semicircle

Given:   \(\theta = \pi\),   \(I\)\( = 2\,A\,,\,\,\, r = \)\(1 × 10^{-2} m\)

\(\vec B_{BC} = \dfrac {10^{-7} × 2 × \pi }{1 × 10^{-2}} \)

\(\vec B_{BC} = 2 \pi × 10^{-5} \) [outward]

image image

So, total magnetic field at O,

\(\vec B = \vec B_{AB}+\vec B_{BC}+\vec B_{CD}\)

\(\vec B = 0 + 2 \pi × 10^{-5}+0\)

\(\vec B =2 \pi × 10^{-5} \,T\) [outward]

image

Calculate the magnetic field at point O due to current carrying element with current \(I\) = 2A as shown in figure.[Given r = 1 cm] [Assume the plane of the paper as x-y plane]

image
A

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

.

B

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

C

\(0\,T\)

D

\(2\pi × 10^{-5}\, T\)

Option D is Correct

Illustration Questions

Two infinite long straight wires carrying current of \(I_1=1\,A\) and \(I_2=2\,A\) respectively, as shown in the figure, are placed at a distance \(\ell = 2 \,cm \) apart. Calculate the magnetic field at point P at the center of two straight wires.  

A \(4×10^{-5}\, T (outwards)\)

B \(6×10^{-5} \,T (outwards)\)

C \(2×10^{-5} \,T (inwards)\)

D \(2×10^{-5}\, T (outwards)\)

×

Magnetic field at P due to \(1A\) of wire ,

\(\vec B_{1A} = \dfrac {\mu_0} {4\pi} \dfrac {I} {d} [sin\theta_1+sin\theta_2]\)

\(\vec B_{1A} = 10^{-7}\dfrac {1} {1×10^{-2}} [sin\dfrac {\pi}{2}+sin \dfrac {\pi}{2}]\)

\(\vec B_{1A}=\) \(2×10^{-5} \,T (inwards)\)

 

image image

Magnetic field at P due to \(2A\) of wire ,

\(\vec B_{2A} = \dfrac {\mu_0} {4\pi} \dfrac {2} {1×10^{-2}} [sin\dfrac {\pi}{2}+sin\dfrac {\pi}{2}]\)

\(\vec B_{2A} =\)\(4×10^{-5} \,T (outwards)\)

image image

Hence, total magnetic field at P wire be 

\(\vec B = \vec B_{1A}+\vec B_{2A}\)

\(\vec B \) = \(2×10^{-5} \,T (inwards)\) + \(4×10^{-5} \,T (outwards)\)

\(\vec B \) = \(2×10^{-5} \,T (outwards)\)

 

image

Two infinite long straight wires carrying current of \(I_1=1\,A\) and \(I_2=2\,A\) respectively, as shown in the figure, are placed at a distance \(\ell = 2 \,cm \) apart. Calculate the magnetic field at point P at the center of two straight wires.  

image
A

\(4×10^{-5}\, T (outwards)\)

.

B

\(6×10^{-5} \,T (outwards)\)

C

\(2×10^{-5} \,T (inwards)\)

D

\(2×10^{-5}\, T (outwards)\)

Option D is Correct

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