Force on a moving charge in a magnetic field

Let us consider a positive charge ‘q’ moving with uniform velocity v. Then the charge ‘q’ will experience a force  which depends upon the following factors i.e. The magnitude of the force is

(i) directly proportional to the moving charge i.e.
\begin{equation}
F \propto q \quad-(1)
\end{equation}

(ii) directly proportional to the applied magnetic field i.e.
\begin{equation}
F \propto B \quad-(2)
\end{equation}

(iii) directly proportional to the sine component of velocity vector i.e.
\begin{equation}
F \propto v \sin \theta \quad-(3)
\end{equation}

Hence from above three equations, we conclude that
\begin{equation}
F \propto q v B \sin \theta \quad \text { or } \quad F=k q v B \sin \theta
\end{equation}

where k = constant whose value is one for each system.

Hence above equation becomes
\begin{equation}
F=q v B \sin \theta \Rightarrow \overrightarrow{\boldsymbol{F}}=q(\vec{v} \times \vec{B})
\end{equation}  -(4)

The direction of this force can be felt by Fleming left hand rule.

Units & Dimensional formula of Magnetic Field :-

A) UNITS :-

(i) In S.I. System, unit of B is Tesla (T) or wb/m²

(ii) In C.G.S. System, unit of B is Gauss (G)

\begin{equation}
1 \text { Tesla }=10^4 \text { Gauss }
\end{equation}

Also, from eqn (4), we know that

\begin{equation}
\begin{aligned}
&F=q v B \sin \theta \Rightarrow B=\frac{F}{q v \sin \theta}\\
&1 \text { Tesla }=\frac{1 \text { Newton }}{1 \text { coulomb } \times 1 \mathrm{~m} / \mathrm{sec}}
\end{aligned}
\end{equation}

B) DIMENSIONAL FORMULA :-

From eqn (5), we can calculate the dimensional formula for B, which is given as

\begin{equation}
B=M L^{\circ} T^{-2} A^{-1}
\end{equation}

BIOT–SAVART’S LAW :-

Let us consider a straight conductor of length ‘l’. Take a small length element P Q = dl on this straight current carrying conductor.

We have to find magnetic field at any point P which is at a distance ‘r’ from the center of small element portion. Let ‘0’ be the angle between r & dl. Now, the small magnetic field (dB) due to a very small… then

\begin{equation}
\oint \vec{B} \cdot d \vec{l}=\oint B d l \cos 0^{\circ}=\oint B d l=B \times 2 \pi l
\end{equation}

Also, according to Ampere’s circuital law

\begin{equation}
\oint \vec{B} \cdot d \vec{l}=\mu_0 I
\end{equation}

If the coil has ‘n’ no. of turns each of length ‘l’, then eqn (2) becomes

\begin{equation}
\oint \vec{H} \cdot d \vec{l}=\mu_0 n I l=\mu_0 n I \times 2 \pi l
\end{equation}

Hence from eqn (1) & (3), we have

\begin{equation}
\begin{gathered}
B \times 2 \pi l=\mu_0 n I \times 2 \pi l \\
B=\mu_0 n I \\
B=\frac{\mu_0 I}{2 \pi l}
\end{gathered}
\end{equation}

This above result may also be written as

\begin{equation}
B=\frac{\mu_0}{4 \pi} \frac{2 I}{r}
\end{equation}

From the above result, one can conclude that the magnitude of magnetic field at every point on every circle of radius ‘r’ with a wire along the axis is same. It means that the magnetic field due to current through infinite straight wire has a cylindrical symmetry

Lorentz force :-

The Lorentz force is the force experienced by a charged particle moving in space where both electric & magnetic field exist is called Lorentz force. Simply, this force is a combination of force due to electric field (Fₑ) & force due to magnetic field .

Force due to electric field :-

As we know that the electric field due to a charged particle carrying +q charge is given by

\begin{equation}
\vec{E}=\frac{\vec{F}}{q} \Rightarrow \vec{F}=q \vec{E}
\end{equation}

Force due to magnetic field :-

As we know that the force acting on a moving charge when placed in magnetic field is given by

\begin{equation}
\begin{aligned}
&F=q v B \sin \theta \Rightarrow\\
&F=q(\vec{V} \times \vec{B})
\end{aligned}
\end{equation}

On Combining (1) & (2), we have the Lorentz force (Fₗ) is

\begin{equation}
F_L=q \vec{E}+q(\vec{V} \times \vec{B}) \Rightarrow F_L=q(\vec{E}+\vec{V} \times \vec{B})
\end{equation}

Which is the required expression for Lorentz force.

Force acting on a current–carrying conductor placed in magnetic field

Proof :-
As we know that the force acting on a moving charge is given by

\begin{equation}
f=q v B \sin \theta \quad-(1)
\end{equation}

As we know that the moving charge produces a current, which is given by

\begin{equation}
I=\frac{q}{t} \Rightarrow q=I t
\end{equation}

Also the charge particle moving in a conductor of length l with a velocity (drift velocity) is given by

\begin{equation}
v=\frac{l}{t} \quad-(3)
\end{equation}

Hence, by using (3) & (2) in (1), we have

\begin{equation}
f=I t \times \frac{l}{t} \times B \sin \theta \Rightarrow F=I l B \sin \theta
\end{equation}

Also, if  is the force on each electron then total force acting on all the electrons is

\begin{equation}
\begin{array}{cc}
\vec{F}=\mathrm{N} \vec{F}=\mathrm{nA} \vec{l}\left(-\mathrm{e}\left[\vec{V}_d \times \vec{B}\right]\right) & {[\text { using } 2 \& 3]} \\
\Rightarrow \vec{F}=-n A \vec{l} e\left(\vec{V}_d \times \vec{B}\right) &
\end{array}
\end{equation}

Also, we know that, the relation between current & drift velocity is

$$
\begin{gathered}
I=\text { Anev }_d \\
\Rightarrow I l=\text { Anev }_d l \quad[\text { multiply both sides }]
\end{gathered}
$$

In vector form

$$
i \vec{l} \times B=- \text { Ane } \vec{V}_d l-(5)
$$

Substitute this value in eqn (4), we have

$$
\vec{F}=I \vec{l} \times \vec{B}
$$

or

$$
F=I l B \sin \theta
$$

Forces between two infinitely long straight IIal conductors :-

Let us consider two parallel infinite straight current carrying conductors PQ & RS.
By using Right hand thumb rule & then by using Fleming left hand rule, we can find the direction of magnetic field & force respectively.

Now, magnetic field intensity B₁ at any point A on conductor RS due to current I₁ passing through PQ is given by

\begin{equation}
B_1=\frac{\mu_0 2 I_1}{4 \pi r}-(1)
\end{equation}

Also, we know that force on a current carrying conductor placed in a magnetic field is given by the relation

\begin{equation}
\begin{aligned}
&F=I_2 B_1\\
&\text { (2)[max. force] }
\end{aligned}
\end{equation}

but this force is due to current I₂ flowing through RS. Hence for a unit length, equation (2) becomes

\begin{equation}
F_2=I_2 B_1
\end{equation} – (3)

Substitute the value of B₁ from eq (1) in eq (3), we have

\begin{equation}
\begin{aligned}
&F_2=\frac{\mu_0 2 I_1 I_2}{4 \pi r}\\
&\text { Similarly, we can find } \mathbf{F}_{\mathbf{1}}=\frac{\mu_0 2 I_1 I_2}{4 \pi r}
\end{aligned}
\end{equation} – (4)