Current I₁ = 2A ; I₂ = 5A
Two wires are separated by a distance a = 1 cm
= 1 x 10^-2m
The force between two parallel wires per unit length
F =?
F = \(\frac { { \mu }_{ 0 } }{ 2\pi } .\frac { { I }_{ 1 }{ I }_{ 2 } }{ a } \)
= 2 x 10^-7 x \(\frac { 2\times 5 }{ 1\times { 10 }^{ -2 } } \)
F = 20 x 10^-5 N
This force is repulsive F = 20 x 10^-5 N.

In the general case, the unit normal vector \(\hat { n } \) and magnetic field \(\vec { B } \) is with an angle 8 as shown in Figure.

(a) The force on section PQ
\(\vec { i } =a\hat { j } \) and \(\vec { B } =B\hat { i } \)
\(\vec { { F }_{ PQ } } =\vec { Il } \times \vec { B } =IaB(\hat { j } \times \hat { j } )=-IaB\hat { k } \)
Since the unit vector normal to the plane \(\hat { n } \) is along the direction of .\(\vec { k } \)
(b) The force on section QR
\(\vec { l } =bcos\left( \frac { \pi }{ 2 } -\theta \right) \hat { i } -sin\left( \frac { \pi }{ 2 } -\theta \right) \hat { k } \)
\(\vec { { F }_{ QR } } =\vec { Il } \times \vec { B } =-IbB\left( \frac { \pi }{ 2 } -\theta \right) \hat { j } \)
\(\vec { { F }_{ QR } } =-IbBcos\theta \hat { j } \)
(c) The force on section RS
\(\vec { l } =a\hat { j } \) and \(\vec { B } =B\hat { i } \)
\(\vec { { F }_{ RS } } =\vec { Il\times \vec { B } } =IaB(\hat { j } \times \hat { j } )=-IaB\hat { k } \)
Since the unit vector normal to the plane is along the direction of \(\hat { k } \).
(d) The force on section SP
\(\vec { l } =bcos\left( \frac { \pi }{ 2 } -\theta \right) \hat { i } +sin\left( \frac { \pi }{ 2 } +\theta \right) \hat { k } \quad \vec { B } =B\hat { i } \)
\(\vec { { F }_{ SP } } =\vec { Il } \times \vec { B } =IbBsin\left( \frac { \pi }{ 2 } -\theta \right) \hat { j } \)
\(\vec { { F }_{ SP } } =-IbBcos\theta \hat { j } \)
The net force on the rectangular loop is
\(\vec { { F }_{ net } } =\vec { { F }_{ PQ } } +\vec { { F }_{ QR } } +\vec { { F }_{ RS } } +\vec { { F }_{ SP } } \)
\({ F }_{ net }=IaB\hat { k } -IbBcos\theta \hat { j } -IaB\hat { k } +IbBcos\theta \hat { j } \)
\(\vec { { F }_{ net } } =\vec { 0 } \)
Hence, the net force on the rectangular loop in this configuration is also zero. Notice that the force on section QR and SP is not zero here. But, they have equal and opposite effects, but we assume that the loop to be rigid, so no deformation. So, no torque was produced by these two sections.
Even though the forces PQ and RS also are equal and opposite, they are not collinear. So these two forces constitute a couple as shown in Figure (a). Hence the net torque produced by these two forces about the axis of the rectangular loop is given by
\(\vec { { \tau }_{ net } } =baBIsin\theta \hat { k } =ABIsin\theta \hat { k } \)

\(\vec { OA } =\frac { b }{ 2 } cos\left( \frac { \pi }{ 2 } -\theta \right) (-\hat { i } )+\frac { b }{ 2 } cos\left( \frac { \pi }{ 2 } -\theta \right) (-\hat { k } )\)
=\(\frac { b }{ 2 } (-sin\theta \hat { i } +cos\theta \hat { k } )\)
\(\vec { OB } =\frac { b }{ 2 } cos\left( \frac { \pi }{ 2 } -\theta \right) (\hat { i } )+\frac { b }{ 2 } cos\left( \frac { \pi }{ 2 } -\theta \right) (\hat { k } )\)
=\(\frac { b }{ 2 } (-sin\theta \hat { i } +cos\theta \hat { k } )\)
\(\vec { OA } \times \vec { { F }_{ PQ } } =\left\{ \frac { b }{ 2 } (-sin\theta \hat { i } +cos\theta \hat { k } \right\} \times \left\{ IaB\hat { k } \right\} \)
= \(\frac { 1 }{ 2 } IabBsin\theta \hat { j } \)
\(\vec { OA } \times \vec { { F }_{ RS } } =\left\{ \frac { b }{ 2 } (sin\theta \hat { i } +cos\theta \hat { k } \right\} \times \left\{ -IaB\hat { k } \right\} \)
= \(\frac { 1 }{ 2 } IabBsin\theta \hat { j } \)
The net torque \(\vec { \tau _{ net } } =IaBsin\theta \hat { j } \)         ..........(1)
Note that the net torque is in the positive y-direction which tends to rotate the loop in a clockwise direction about the y axis. If the current is passed in the other way (P⟶S⟶R⟶Q⟶P), then total torque will point in the negative y-direction which tends to rotate the loop in an anticlockwise direction about the y-axis.
Another important point is to note that the torque is less in this case compared to the earlier case (where the \(\hat { n } \) is perpendicular to the magnetic field \(\vec { B } \)). It is because the perpendicular distance is reduced between the forces \(\vec { { F }_{ PQ } } \) and \(\vec { { F }_{ RS } } \) in this case.
The equation (1) can also be rewritten in terms of magnetic dipole moment \(\vec { { p }_{ m } } =I\vec { A } =Iab\hat { n } \)
\(\vec { \tau _{ net } } =\vec { p } \times \vec { B } \).

(i) Consider a charged particle of charge 'q' having mass m enters into a region of a uniform magnetic field \(\vec { B } \) with velocity \(\vec { v } \).
(ii) Such that velocity is perpendicular to the magnetic field and velocity \(\vec{v}\).
(iii) The charged particle moves in a circular orbit.
(iv) Lorentz force.

\(\vec { F } =q(\vec { v } \times \vec { B } )\)
In magnitude F = qVB
(v) This Lorentz force acts as centripetal force for the particle to execute circular motion. Therefore,
qvB = m\(\frac { { v }^{ 2 } }{ r } \)
The radius of the circular path is
r = \(\frac { mv }{ qB } =\frac { p }{ qB } \)               ..........(1)
(vi) where p = mv is the magnitude of the linear momentum of the particle. Let T be the time taken by the particle to finish one complete circular motion, then
T = \(\frac { 2\pi r }{ v } \)                  .............(2)
Hence substituting (1) in (2), we get,
T = \(\frac { 2\pi m }{ qB } \)        .............(3)
(vii) Equation (3) is called the cyclotron period. The reciprocal of time period is the frequency f, which is
f = \(\frac { 1 }{ T } \)
f = \(\frac { qB }{ 2\pi m } \)                  ...........(4)
In terms of angular frequency ω,
ω = 2πf = \(\frac { q }{ m } \)B          ...........(5)
(viii) Equations (4) and equation (5) are called cyclotron frequency or gyrofrequency.
(ix) Time period and frequency depend only on charge-to-mass ratio (specific charge) and independent of velocity or radius.

(i) Electron revolves around a nucleus in a circular orbit of radius R.
(ii) Circulating electron is like a current in a circular loop.

\(\vec { { \mu }_{ L } } =I\vec { A } \) ......(1)
In magnitude,
μ_L = IA
If T is the time period of an electron, the current due to the circular motion of the electron is
I = \(\frac { -e }{ T } \) ....(2)
where -e is the charge of an electron. If R is the radius of the circular orbit and v is the velocity of the electron in the circular orbit, then
T = \(\frac { 2\pi R }{ v } \)   ..(3)
Using equation (2) and equation (3) in equation (1), we get
μ_L = \(\frac { e }{ \frac { 2\pi R }{ v } } \pi { R }^{ 2 }=\frac { evR }{ 2 } \)   ....(4)
where A = πR² is the area of the circular loop. By definition, the angular momentum of the electron about O is
\(\vec { L } =\vec { r } \times \vec { p } \)
In magnitude
L = Rp = mvR  ...(5)
Using equation (4) and equation (5), we get
\(\frac { { \mu }_{ L } }{ L } =-\frac { \frac { evR }{ 2 } }{ mvR } =\frac { e }{ 2m } \Rightarrow \vec { { \mu }_{ L } } =\frac { e }{ 2m } \vec { L } \) ....(6)
The negative sign indicates that the magnetic moment and angular momentum are in opposite directions.
In magnitude
\(\frac{\mu_L}{L}=\frac{e}{2 m}=\frac{1.60 \times 10^{-19}}{2 \times 9.11 \times 10^{-31}}=0.0878 \times 10^{12} \mathrm{C} \mathrm{kg}^{-1}\)
\(\frac{\mu_L}{L}=8.78 \times 10^{10} \mathrm{C} \mathrm{kg}^{-1}=\text { constant }\)
The ratio \(\frac{\mu_L}{L}\) is aconstant known as Gyro-magnetic ratio \(\left(\frac{e}{2 m}\right)\).
According to Bohr quantization
\(\mathrm{L}=\mathrm{nh} / 2 \pi\)
\(\mu_L=\left(\frac{e}{2 m}\right) \mathrm{L}=\frac{\pi e h}{4 \pi m} \)
On substiting known values
\(\mu_L=9.27 \times 10^{-24} \mathrm{~A} \mathrm{~m}^2\)
The minimum magnetic moment can be obtained by substituting n = 1
\(\mu_2=9.27 \times 10^{-24} \mathrm{~A} \mathrm{~m}^2=9.27 \times 10^{-24} \mathrm{~J} \mathrm{~T}^{-1}\)
where, \(\mu_B=\frac{c h}{4 \pi m}=9.27 \times 10^{-24} \mathrm{~A} \mathrm{~m}^2\) is called Bohr magneton. which is used to measure atomic magnetic moments.

When a bar magnet (magnetic dipole) of dipole moment \(\vec { { p }_{ m } } \) is held at an angle θ with the direction of a uniform magnetic field \(\vec { { B } } \), as shown in Figure the magnitude of the torque acting on the dipole is

\(|\vec { \tau _{ B } } |=|\vec { { p }_{ m } } ||\vec { B } |sin\theta \)
If the dipole is rotated through a very small angular displacement dθ against the torque ፒ_B at constant angular velocity, then the work done by external torque \((\vec { { \tau }_{ ext } } )\) for this small angular displacement is given by
dW = \(|\vec { { { \tau }_{ ext } } } |\) dθ
Since the bar magnet to be moved at constant angular velocity, it implies \(|\vec { { \tau }_{ B } } |=|\vec { \tau _{ ext } } |\)
dW = P_mB sinθ dθ
Total work done in rotating the dipole from θ' to θ is
W =\(\int _{ \theta ' }^{ \theta }{ \tau d\theta } =\int _{ \theta ' }^{ \theta }{ p_{ m } } Bsin\theta d\theta ={ p }_{ m }B[-cos\theta d\theta ]_{ \theta ' }^{ \theta }\)
W = p_mB(cosθ - cosθ')
This work done is stored as potential energy in bar magnet at an angle θ when it is rotated from θ' to θ and it can be written as
U = p_mB(cosθ - cosθ')           .........(1)
In fact, equation (1) gives the difference in potential energy between the angular positions θ' and θ. We can choose the reference point θ' = 90°, so that second term in the equation becomes zero and the equation (1) can be written as
U = -p_mB(cosθ)              ............(2)
The potential energy stored in a bar magnet in a uniform magnetic field is given by
U = -\(\vec { { p }_{ m } } .\vec { B } \)                     ............(3)
Case 1
(i) If θ = 0°, then
U = P_mB (cos0⁰) = - P_mB
(ii) If θ = 180°, then
U = P_mB (cos 180°) = p_mB
We can infer from the above two results, the potential energy of the bar magnet is minimum when it is aligned along the external magnetic field and maximum when the bar magnet is aligned anti-parallel to an external magnetic field.