Magnetic Effects Of Current and Magnetism

The magnetic effect of current is the phenomenon where an electric current creates a magnetic field around a conductor. This was first observed by Hans Christian Oersted. Laws like Biot-Savart and Ampere’s Circuital Law help us understand how these fields are formed. Magnetic field lines show the direction of the field around the wire. Devices like solenoids and toroids use this effect to generate uniform magnetic fields. Depending on how materials respond to these fields, they are classified as ferromagnetic, paramagnetic, or diamagnetic.

1.0Concept of Field and Oersted Experiment

A region around any physical quantity where another similar physical quantity experiences force or torque is called a field.

2.0Oersted Experiment

When the direction of current in the conductor is reversed then deflection of the magnetic needle is also reversed. Increasing current or moving the needle closer increases its deflection.

3.0Biot Savart Law

The magnetic field at a point is directly proportional to the current, element length, and sine of the angle, and inversely proportional to the square of the distance.

$dB=\frac{{\mu }_0}{4\pi }\frac{IdlSin\theta }{r^2}$

${\mu }_0=4\pi \times 10^{-7}Tm{a}^{-1}$

${\mu }_0$: Permeability of free space

$d\vec{B}=\frac{{\mu }_0}{4\pi }\frac{I(d\vec{l}\times \vec{r})}{r^3}\Rightarrow \text{Vector Form}$

$\vec{B}=\int d\vec{B}=\frac{{\mu }_0}{4\pi }\int \frac{I(d\vec{l}\times \vec{r})}{r^3}\text{(Integral Form)}$

4.0Application of Biot-Savart Law

1. Magnetic Field due to Thin Wire of Finite Length

$B=\frac{{\mu }_{0}I}{4\pi a}[sin{\varphi }_1+Sin{\varphi }_2]\text{inwards}$

2. Magnetic Field due to Infinite Straight Wire

$B=\frac{{\mu }_{0}I}{2\pi a}$

3. Magnetic Field due to Semi Infinite Straight Wire

Case-1

${\varphi }_1\to 90^{o}or{\varphi }_2\to 0°$

$B={\frac{{\mu }_{0}I}{4\pi a}}^{-\otimes }$

Case-2.

${\varphi }_1\to 90^{o}or{\varphi }_2\to \theta$

$B=\frac{{\mu }_{0}I}{4\pi a}(1+sin\theta {)}^{-\otimes }$

Case-3.

${\varphi }_1\to 90^{o}or{\varphi }_2\to -\theta$

$B=\frac{{\mu }_{0}I}{4\pi a}(1-sin\theta {)}^{-\otimes }$

Case-4.

${\varphi }_1\to -\theta or{\varphi }_2\to 90^0$

$B=\frac{{\mu }_{0}I}{4\pi a}(1-sin\theta {)}^{-\otimes }$

5.0Magnetic Field due to Circular Loop

Direction: Can be obtained by right hand thumb rule -Curl the right-hand fingers along the current; the thumb shows the magnetic field direction.

1. Magnetic Field due to Circular Coil

For N loop- $B_O=\frac{{\mu }_{0}NI}{2R}$

2. Magnetic Field due to Circular Arc

$B=\frac{{\mu }_0}{4\pi }\frac{I\alpha }{R}$

3. Wire of circuit loop is uniform

$\frac{B_{abc}}{B_{adc}}=(\frac{l_2}{l_1})(\frac{l_1}{l_2})=\frac{1}{1}$

The magnetic field produced by both the arc is at the centre of the circuit loop is equal in intensity and in reverse direction. So $(B_O{)}_{net}=0$ (always and it is free from angle of connection of terminals)

4. Wire of circuit loop is non-uniform

Case-1. Thickness-Same, Material - different	Case-2. Thickness-Different, Material - Same
$(B_O{)}_{net}=\frac{{\mu }_O}{4r}(I_1-I_2)$	$(B_O{)}_{net}=\frac{{\mu }_O}{4r}(I_1-I_2)$

Note: When current divides in any symmetrical planar loop made with uniform wire, then the magnetic field at the centre due to this loop is ZERO.

5. Magnetic Field at the Axis of Circular loop

$B_P=\frac{{\mu }_{0}I{R}^2}{2(R^2+x^2{)}^{\frac{3}{2}}}=\frac{B_o}{(1+\frac{x^2}{R^2}{)}^{\frac{3}{2}}}$

Case-1: At very large distance from centre, $B_P=\frac{{\mu }_{O}I{R}^2}{2x^3}$

Case-2: Near centre of the ring $B_P=\frac{B_o}{(1+\frac{x^2}{R^2}{)}^{\frac{3}{2}}}$

6.0Ampere’s Circuital Law and it's Utilities

Ampere’s Circuital Law: Line integral of magnetic field along any closed loop is equal to o times the net current crossing the surface bounded by the loop.

$\oint \vec{B}.\overrightarrow{dl}={\mu }_0{I}_{enc}$

Note:

1. It is applicable only for steady/constant current

2. It can be applied for any distribution of current but, it is applied for symmetric distribution for calculation purposes.

1. Magnetic Field due to Long Thin wire

$B=\frac{{\mu }_{0}I}{2\pi r}$

2. Magnetic Field due to Long Thick wire

$B_{Out}=\frac{{\mu }_{0}I}{2\pi r}$

$B_{sur}=\frac{{\mu }_{0}I}{2\pi R}$

$B_{in}=\frac{{\mu }_{0}I}{2\pi R^2}=\frac{{\mu }_{0}Jr}{2}$

The magnetic field is maximum at the surface of the wire.

3. Magnetic Field due to Long Hollow Cylindrical Conductor

7.0Magnetic field due to Solenoid

It is a wire wound into a helix with insulated turns. The magnetic field inside is strong and aligned along the axis, while outside it is nearly zero. The field direction inside is determined using the right-hand thumb rule.

Magnetic Field Inside a Long Solenoid

$B={\mu }_{O}nI={\mu }_O(\frac{N}{l})I$

M.F. outside the solenoid $B=0$

M.F. at the edges/end points $B=\frac{{\mu }_{O}nI}{2}$

Key Note: A solenoid creates a magnetic field similar to that of a bar magnet, acting as a magnetic dipole and serving as an electromagnet in various devices.

8.0Magnetic field due to Toroid

A toroid can be considered as a ring-shaped closed solenoid also called endless solenoid. Toroid is a solenoid bent in ring shape.

Point A: Magnetic field in the empty space surrounded by toroid $(r<R_1),B=0$

Point B: Magnetic field outside the toroid $(r>R_2),B=0$

Point C: Inside the Toroid $(R_1\le r\le R_2)B={\mu }_{O}nI$

$n=\frac{N}{2\pi r_m}$$[r_m=\frac{R_1+R_2}{2}$$=\text{Mean Radius of Toroid}]$

9.0Motion of Charge in Magnetic Field

Magnetic Force on Moving Charge

$\vec{F}=\vec{q}(\vec{v}\times \vec{B})=qvBsin\theta$

Magnetic force depends on angle between $\vec{v}$  and  $\vec{B}$

Case-1: θ = 0° or 180°	Case-2: θ = 90°
F=0	$F_{max}=qvB$

Note: Direction of force can be identified by right hand thumb rule or right hand palm rule.

Motion of Charged Particle in an Unvarying Magnetic Field

Case 1:

1. $\theta$=0°

$F=qvBsin0°=0$                                                    

Straight Line Path

2. $\theta =180°$

$F=qvBsin1850°=0$

Straight Line Path

Case 2: Motion of charge particle in uniform transverse magnetic field

$\text{Radius of Curvature},R=\frac{mv}{qB}(\because p=mv)$

$R=\frac{p}{qB}=\frac{\sqrt{2mK}}{qB}K:\text{Kinetic Energy}$

$\text{K.E, }K=q{V}_{\text{acc}}$

$R=\frac{\sqrt{2mq{V}_{\text{acc}}}}{qB}\Rightarrow R=\frac{1}{B}\sqrt{\frac{2m{V}_{\text{acc}}}{q}}$

$\text{Time period (T)}:T=\frac{2\pi m}{qB}$

$\text{Frequency},f=\frac{1}{T}=\frac{qB}{2\pi m}$

$\text{Angular frequency},\omega =\frac{qB}{m}$

Case-3.Motion of Charge Particle in Oblique Magnetic Field (  0°,90°,180°)

Radius of Circular Path, $R=\frac{mvsin\theta }{qB}$

Time period of circular motion, $T=\frac{2\pi m}{qB}$

Pitch of helix (p):The linear distance travelled by the charge particle in one revolution or in one time period along the external magnetic field direction is called 'pitch of helix'. $T=\frac{2mvcos\theta }{qB}$

Lorentz Force: When a charge moves in an electric field (E) and magnetic field (B), both forces act on it. The combined effect is called the Lorentz force.

$\overrightarrow{F_{net}}=q(\vec{v}\times \vec{B})+(q\vec{E})$

10.0Magnetic Force on a Current Carrying Wire

$\overrightarrow{dF}=I(\overrightarrow{dl}\times \vec{B})$

$\vec{F}=I(\overrightarrow{L_{eff}}\times \vec{B})L_{eff}$ is the displacement vector from starting point of current to end

point of current.

11.0Magnetic Interaction Between Two Parallel Current Carrying Wires

Magnetic force per unit length of each conductor is, $\frac{d{F}_{12}}{dl}=\frac{d{F}_{21}}{dl}=\frac{{\mu }_0{I}_1{I}_2}{2\pi d}$

12.0Definition of Ampere

• Magnetic force/unit length for both infinite length conductor gives as  

$f=\frac{{\mu }_0{I}_1{I}_2}{2\pi d}=\frac{(4\pi \times 10^{-7})((1)1)}{2\pi (1)}=2\times 10^{-7}N/m$

• 'Ampere is the current which, when passed through each of two parallel infinite long straight conductors placed in free space at a distance of 1 m from each other, produces between them a force of 2 × 10^–7 N/m

13.0Bar Magnet and Magnetic Dipole

Bar Magnet: It is made up of iron, steel or any other ferromagnetic substance or ferromagnetic composite, having permanent magnetic properties. Two poles are present in a Bar magnet, North Pole and South Pole.

Magnetic Dipole: A magnetic dipole consists of a duo of magnetic poles of equal and opposite strength separated by a small distance.

Example: Magnetic needle, Bar Magnet, Current carrying coil/solenoid etc.

$\vec{M}=m\vec{l}$

• Vector quantity
• Direction: South → North
•  Unit: $A-m^2$ or J/T

Magnetic Moment of Current Carrying Coil (Loop)

Magnetic Moment $\vec{M}=NI\vec{A}$

14.0Torque and Force on Magnetic Dipole

1. Bar Magnet

$\vec{\tau }=\vec{M}\times \vec{B}=MBsin\theta$

$\tau =MBsin\theta$

Case-1. If $\theta =90°\Rightarrow \tau =MB(maximum)$

Case-2.If $\theta =0°or180°\Rightarrow \tau =0(minimum)$

2. Coil or Loop

$\tau =NIABsin\theta$

Case-1. If $\theta =90°\Rightarrow \tau =NIAB(maximum)$

Case-2.If $\theta =0°or180°\Rightarrow \tau =0(minimum)$

15.0Moving Coil Galvanometer

The galvanometer has a coil with many turns, free to rotate in a uniform radial magnetic field. A soft iron core strengthens and radializes the field, while a spiral spring resists the coil's rotation.

• A current bearing coil in a magnetic field experiences a torque.  $\tau =NIABsin\theta$
• The spring S provides a counter torque C that balances the magnetic torque, ${\tau }^{\prime }=C\varphi$
• In equilibrium,$C\varphi =NIAB\Rightarrow I=\frac{C}{NAB}\varphi$, $I\propto \varphi$

It means the deflection produced is proportional to the current flowing through the galvanometer.

16.0Potential Energy of Magnetic Dipole

Work Done in Rotating the Coil in Uniform Magnetic Field

$W=MB(cos{\theta }_1-cos{\theta }_2)$

Potential Energy of the Coil in Uniform Magnetic Field

$U=-MBcos\theta$

$U=-\vec{M}.\vec{B}$

Case-1	Case-2	Case-3
When $\vec{M}and\vec{B}$ are parallel $\theta$= 0° $\tau$ = 0 $U_{min}=-MB$	When $\vec{M}and\vec{B}$ are perpendicular $\theta =90°$ $\tau =MB$ U = 0	When  $\vec{M}and\vec{B}$ are anti- parallel $\theta =180°$ $\tau =0$ $U_{max}=MB$

17.0Atomic Magnetism

18.0Angular SHM of Magnetic Dipole

$T=2\pi \sqrt{\frac{I}{MB}}$                 I:Moment of Inertia =$\frac{M{l}^2}{12}$

19.0Terminology Used In Magnetism

1. Magnetizing field or Magnetic Intensity$(\vec{H})$: Field in which a material is placed for magnetization.         

$\vec{H}=\frac{\overrightarrow{B_0}}{{\mu }_O}=A/m$

2. Intensity of magnetization ($\vec{I}$): The intensity of magnetisation is the induced dipole moment per unit volume when a magnetic material is placed in a magnetising field.

$\vec{I}=\frac{\vec{M}}{V}\Rightarrow A/m$

3. Magnetic susceptibility (${\chi }_m$): It is a scalar with no units & dimensions. Physically it represents the ease with which a magnetic material can be magnetised.

${\chi }_m=\frac{I}{H}$

4. Magnetic Permeability$(\mu )$

$\mu =\frac{B}{H}=\frac{Totalmagneticfieldinsideamaterial}{MagnetizingField}=H/m$

5. Relative Permeability$({\mu }_r)$

${\mu }_r=\frac{\mu }{{\mu }_o}$

It has no units and dimensions.

Relation between $\mu$ and ${\chi }_m$

${\mu }_r=1+{\chi }_m$

20.0Magnetic Materials

21.0Sample Questions on Magnetic Effects of Current and Magnetism 

Q-1. Calculate magnetic field at P

Solution:

Here

$I\overrightarrow{dl}||\vec{r}$ So, $Sin\theta =0$

Hence, the magnetic field at P is zero.

Q-2. If point ‘P’ lies out-side the line of wire then magnetic field at point ‘P’ will be :

Solution:

${\varphi }_1=(90^o-{\alpha }_1){\varphi }_2=(90^o-{\alpha }_2)$

$B_P=\frac{{\mu }_{0}I}{4\pi d}[sin(90^o-{\alpha }_1)-sin(90^o-{\alpha }_2)]=\frac{{\mu }_{0}I}{4\pi d}(cos{\alpha }_1-cos{\alpha }_2)$

Q-3. Circular loop is made by a wire of length 7.5m. If current 5A is flowing in the loop, then find the magnetic field at the centre.

Solution:

$2\pi R=7.5\Rightarrow R=\frac{7.5}{2\pi }\Rightarrow B_c=\frac{{\mu }_{0}I}{2R}$$\Rightarrow \frac{4\pi \times 10^{-7}\times 5}{2\times \frac{7.5}{2\pi }}=\frac{40{\pi }^2\times 10^{-7}}{15}T$

Q-4. Two symmetrical current carrying rings are placed perpendicular to each other with a common centre. If magnetic field at the centre due to one coil is B, then find a net magnetic field.

Solution:

$B_H=\frac{{\mu }_{0}I}{2R}=B\hat{j}$

$B_V=\frac{{\mu }_{0}I}{2R}=B\hat{i}$

$B_{\text{net}}=\sqrt{B_H^2+B_V^2}\Rightarrow B\sqrt{2}$

Q-5.A hollow cylindrical wire carries a current 5A, having inner and outer radii 'R' and 2R respectively. The magnetic field at a point which is 3R/2 distance away from its axis is (R = 5m).

Solution:

Field inside cross-section of conductor

$B_P=\frac{{\mu }_{0}I}{2\pi r}\left(\frac{r^2-a^2}{b^2-a^2}\right)$

$[a=R,b=2R,r=\frac{3R}{2}]$

$B_P=\frac{{\mu }_{0}I}{2\pi \left(\frac{3R}{2}\right)}\left(\frac{{\left(\frac{3R}{2}\right)}^2-R^2}{(2R{)}^2-R^2}\right)=\frac{5}{36}\cdot \frac{{\mu }_{0}I}{\pi R}=\frac{5}{36}\cdot \frac{5\times {\mu }_0}{5\times \pi }=\frac{5}{9}\times 10^{-7}T$