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

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

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

#### 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 Position vector $$|\vec r|$$

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

$$\vec r = +3 \hat i + 4 \hat j$$

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

$$\hat r = \dfrac {\vec r} {|\vec r|}=\dfrac {3\hat i + 4 \hat j} {5}$$ Magnetic field at origin,

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

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

$$0 \;T$$

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.

$$\mu_0 = 4\pi×10^{-7}$$ $$T m/A$$

#### 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

$$d\vec B=\dfrac {\mu_0}{4\pi} I \dfrac {d\vec l × \vec r} {r^3}$$ Since, the distance r is much larger than the length of element of rod.

So,

$$d\vec B=\dfrac {\mu_0}{4\pi} I \dfrac {d l\, sin\theta} {r^2}$$ Given: $$I = 4\,A\,,$$   $$d\ell$$$$=5\,cm\,,\,\,\,\,\, r = 2\, m\,,$$  $$\theta=37°$$

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

$$d\vec B = 3×10^{-9}\,$$$$T$$ 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 $$-\hat k$$. ### 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

$$4×10^{-8} (\hat k)\,T$$

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 $$\theta$$,

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

#### 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 = $$\dfrac {\mu_0 IR^2}{2(R^2+x^2)^{3/2}}$$ Given: $$R = 3 \,cm$$ ,   $$I$$$$= 2\,A \,\,\,\,,\,\,x = 4 \,cm$$

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

$$B=2.88\,\pi × 10^{-6} \ T$$ ### 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

$$2.3 \,\pi× 10^{9} \, T$$

Option C is Correct

#### 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. 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, $$\vec B_{AB} = 0$$  Magnetic field at O due to CD ,

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

So, $$\vec B_{CD} = 0$$  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]  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] ### 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

$$2\pi × 10^{-5}\, T$$

Option D 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 $$\vec B = -\dfrac {\mu_0I} {4\pi d}$$$$[sin\theta]^{-\theta_2}_{\theta_1}$$

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

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

$$\vec B = \dfrac {\mu_0 I} {4\pi d} [sin\theta_1+sin\theta_2]$$ Given : $$\text I$$$$= 2\,A\,,\,\,\,d = 2\,m\,,$$  $$\theta_1=30°\,,$$  $$\theta_2=53°$$

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

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

$$\vec B = 13×10^{-8}\, T$$ Using right hand thumb rule, direction of magnetic field is outward. ### 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

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)

#### 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, $$\theta = \dfrac {\pi} {2}$$ Given : $$R = 2 \,cm\,,$$   $$\theta = \dfrac {\pi} {2}\,,$$    $$I$$$$= 2\,A$$

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

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

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

$$0$$

Option C is Correct

#### 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. 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]  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] Hence, net magnetic field at O due to loop ,

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

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

$$2\sqrt 2 × 10^{-5}\, T$$

Option C is Correct

#### 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 ,

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

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

$$\vec B_{1A}=$$ $$2×10^{-5} \,T (inwards)$$  Magnetic field at P due to $$2A$$ of wire ,

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

$$\vec B_{2A} =$$$$4×10^{-5} \,T (outwards)$$  Hence, total magnetic field at P wire be

$$\vec B = \vec B_{1A}+\vec B_{2A}$$

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

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

$$2×10^{-5}\, T (outwards)$$

Option D is Correct