NEET Physics Moving Charges and Magnetism Notes

NEET Physics Moving Charges and Magnetism Notes

Moving Charges and Magnetism

Moving Charges

The force experienced by a charged particle when it is moving in a uniform magnetic field is

1. Directly proportional to the strength of the magnetic field. i.e., \(F \propto B\)
2. Directly proportional to the magnitude of charge. i.e., \(F \propto q\)
3. Directly proportional to the component of velocity in a direction perpendicular to the direction of the magnetic field. i.e., \(F \propto v\) sinθ

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∴\(\begin{aligned}
& \mathrm{F} \propto \mathrm{qvB} \sin \theta \\
& \mathrm{F}=\mathrm{kqvB} \sin \theta
\end{aligned}\)

In SI the value of k = 1

\(\mathrm{F}=\mathrm{qvB} \sin \theta\)

In vector form, \(\overrightarrow{\mathrm{F}}=\mathrm{q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}})\)

Case 1:

The magnetic field on q is zero if it is at rest (since v=0)

Case 2:

If angle between \(\overrightarrow{\mathrm{v}} \text { and } \overrightarrow{\mathrm{B}} \text { is } 0^{\circ} \text { or } 180^{\circ}\) is 0º or 180º

Then, F = 0 (∵sin 0º = 0 and sin 180º = 0)

Case 3:

When the charged particle moves perpendicular to the direction of magnetic field, then,

F = qvB(1) (∵sin 90º = 1)

\(\mathrm{F}_{\max }=\mathrm{qvB}\)

SI unit of the magnetic field is tesla (T)

1 tesla (T) = 1 Weber meter^-2 (Wbm^-2)

Note:

The magnetic field is commonly expressed in gauss (G).

1 gauss (G) = 10^-4 tesla(T)

1G = 10^-4 T

Lorentz Force

If a charged particle moves in a region where both electric and magnetic fields are present, then the force experienced by the charged particle is called Lorentz force.

\(\begin{aligned}
& \vec{F}_L=\vec{F}_E+\vec{F}_B \\
& \vec{F}_L=q \vec{E}+q(\vec{v} \times \vec{B})=q[\vec{E}+(\vec{v} \times \vec{B})]
\end{aligned}\)

Motion of a charged particle inside a uniform magnetic field

1. When is perpendicular to \(\overrightarrow{\mathbf{B}}\):

Using Flemings left hand rule it can be shown that the force acting on the charged particle is centripetal.

i.e., the necessary centripetal force for the revolution is given by magnetic force.

Moving Charges And Magnetism

Where ‘r’ is the radius of the circular path.

The time period of revolution of a charged particle is given by,

\(\begin{aligned}
&\mathrm{T}=\frac{\text { circumference }}{\text { speed }}=\frac{2 \pi \mathrm{r}}{\mathrm{v}}=\frac{2 \pi}{\mathrm{v}}\left(\frac{\mathrm{mv}}{\mathrm{qB}}\right)\\
&\mathrm{T}=\frac{2 \pi \mathrm{m}}{\mathrm{qB}}
\end{aligned}\)

The frequency of revolution is given by

\(\mathrm{f}=\frac{\mathrm{qB}}{2 \pi \mathrm{m}}\)

Note:

Time period and frequency are independent of velocity of the charged particle and radius of the circular path.

If the velocity of the charged particle increases, then the radius of the path also increases, without any change in time period, and frequency.

2. When and are inclined to each other, the charged particle moves in a helical path:

The distance traveled by the charged particle in the direction of the magnetic field in a time equal to the time period is called pitch of the helical path.

\(\text { pitch }=v \cos \theta \times \frac{2 \pi m}{q B}\)

Cyclotron

It is a device used to accelerate heavy-charged particles.

Maximum velocity of the charged particle in a cyclotron is given by,

\(\mathrm{v}_{\max }=\frac{\mathrm{qBR}}{\mathrm{m}}\)

Where ‘R’ is the radius of the dees.

The maximum kinetic energy gained is given by,

\(\begin{aligned}
& \mathrm{E}_{\max }=\frac{1}{2} \mathrm{mv}_{\max }^2=\frac{1}{2} \mathrm{~m}\left(\frac{\mathrm{qBR}}{\mathrm{m}}\right)^2 \\
& \mathrm{E}_{\max }=\frac{1}{2} \frac{\mathrm{q}^2 \mathrm{~B}^2 \mathrm{R}^2}{\mathrm{~m}}
\end{aligned}\)

Note:

Cyclotron is not suitable for accelerating electrons.

Since the mass of the electron is very small its speed increases rapidly and starts moving at a relativistic speed.

The magnetic field at a point due to a current-carrying conductor can be calculated using Biot–Savart’s law.

According to Biot–Savart’s law the magnetic field dB produced by a small conductor of length d and carrying current I is given by,

\(\mathrm{dB}=\frac{\mu_0}{4 \pi} \frac{\mathrm{Id} \ell \sin \theta}{\mathrm{r}^2}\)

Where ‘r’ is the distance between the current element and the point where we want to calculate the magnetic field and θ is the angle between the direction of the current element and the line joining the current element and the point of observation.

The magnetic field at a point on the axis of a circular loop carrying current

Where N is the number of turns in the coil.

x is the distance between the point of observation and the center of the loop.

Case 1 :

Magnetic field at the center of the loop (x = 0)

\(\begin{aligned}
& \mathrm{B}=\frac{\mu_0}{4 \pi} \frac{2 \pi \mathrm{NI}}{\mathrm{R}} \\
& \mathrm{B}=\frac{\mu_0 \mathrm{NI}}{2 \mathrm{R}}
\end{aligned}\) \(\text { If } \mathrm{N}=1 \text {, then, } \mathrm{B}=\frac{\mu_0 I}{2 \mathrm{R}}\)

Case 2:

When x >> R,

\(\begin{aligned}
& \mathrm{B}=\frac{\mu_0}{4 \pi} \frac{2 \pi \mathrm{NIR}^2}{\mathrm{x}^3} \\
& \mathrm{~B}=\frac{\mu_0}{4 \pi} \frac{2 \mathrm{~m}}{\mathrm{x}^3} \quad(∵\mathrm{m}=\mathrm{NIA})
\end{aligned}\)

Where ‘m’ is known as magnetic moment of the coil.

Ampere’s Circuital Law

It states that “the line integral of magnetic field around a closed loop enclosing an arbitrary area is equal to times the total current passing through the area”.

\(\text { Mathematically, } \oint \overrightarrow{\mathrm{B}} \cdot \mathrm{d} \vec{\ell}=\mu_0 \mathrm{I}\)

Magnetic field due to a long current carrying wire is given by,

Where ‘r’ is the perpendicular distance between the point of observation and the wire.

Magnetic field due to a current carrying wire of finite length is given by

Magnetic Field Due To Moving Charge

Where, φ1and φ2 are the angles, which the lines joining the observation point with the two ends of the conductor make with the normal to the conductor from the observation point.

The magnetic field due to a long straight solenoid is given by,

\(\mathrm{B}=\mu_0 \mathrm{nI}\)

Where ‘n’ is the number of turns per unit length.

The magnetic field produced by a toroid is given by,

\(\mathrm{B}=\mu_0 \mathrm{nI}\)

The magnetic field due to a circular arc of wire subtending an angle θ at the center

If (for 1 complete loop)

\(\mathrm{B}=\frac{\mu_0 \mathrm{I}}{2 \mathrm{r}}\)

Magnetic force on a current-carrying conductor is given by,

\(\mathrm{F}=\mathrm{BI} l \sin \theta\)

In vector form, \(\overrightarrow{\mathrm{F}}=\mathrm{I}(\vec{\ell} \times \overrightarrow{\mathrm{B}})\)

The force between two infinitely long parallel current-carrying conductors is:

\(\mathrm{F}=\frac{\mu_0}{4 \pi} \frac{2 \mathrm{I}_1 \mathrm{I}_2}{\mathrm{r}}\)

∴ The force per unit length is given by,

\(\mathrm{f}=\frac{\mathrm{F}}{l}=\frac{\mu_0}{4 \pi} \frac{2 \mathrm{I}_1 \mathrm{I}_2}{\mathrm{r}}\)

Definition of Ampere

Let I1 = I2 = 1A, r = 1m, then

\(\mathrm{f}=10^{-7} \times \frac{2 \times 1 \times 1}{\mathrm{r}}=2 \times 10^{-7} \mathrm{~N}\)

Thus, one ampere is that current, which when flowing through each of the two parallel conductors of infinite length and placed in free space at a distance of 1 metre from each other, produces between them a force of newton per meter of their lengths.

The torque acting on a current loop placed in a magnetic field:

\(\tau=\text { BIA } \sin \theta\)

If there are N loops, then

\(\tau=\text { NBLA } \sin \theta
Or
\tau=\mathrm{mB} \sin \theta \quad(∵\mathrm{m}=\mathrm{NIA})\)

Mathematically, \(\vec{\tau}=\vec{m} \times \vec{B}\)

Figure of merit of a galvanometer is the current required to produce unit deflection in the galvanometer

The current sensitivity of a galvanometer is given by,

\(I_S=\frac{\theta}{I}=\frac{N B A}{K}\)

The voltage sensitivity of a galvanometer is given by,

\(\mathrm{V}_{\mathrm{S}}=\frac{\theta}{\mathrm{V}}=\frac{\theta}{\mathrm{IR}}=\frac{\mathrm{NBA}}{\mathrm{KR}}\)

Conversion of galvanometer into an ammeter

A galvanometer can be converted into an ammeter by connecting a low resistance called a shunt in parallel with the galvanometer coil.

\(S=\frac{I_g G}{I-I_g}\)

The resistance of the ideal ammeter is zero.

Conversion of a galvanometer into a voltmeter

This can be done by connecting a high-resistance R in series with the galvanometer.

\(R=\frac{V}{I_g}-G\)

Resistance of a ideal voltmeter is infinity.