Hysteresis is the phenomenon of lagging of magnetic induction behind the magnetising field.

Ampere's circuital law states that the line integral of magnetic field over a closed loop is μ₀, times net current enclosed by the loop.
\(\oint _{ c }^{ }{ \vec { B } \vec { dl } } \) = μ₀I _enclosed.

Electric field, E = 6.0 x 10⁶ N C^-1 and magnetic field, B = 0.83 T.
Then.
\(v=\frac { E }{ B } =\frac { { 6.0\times 10 }^{ 6 } }{ 0.83 } =7.23\times { 10 }^{ 6 }{ ms }^{ -1 }\)
When an electron goes with this velocity, it shows null deflection. Since the accelerating potential is 200 V, the electron acquires kinetic energy because of this accelerating potential. Hence,
\(\frac { 1 }{ 2 } mv^{ 2 }=eV \)
\(v=\sqrt { \frac { 2eV }{ m } }\)
Since the mass of the electron, m = 9.1 x 10⁻³¹kg and charge of an electron, \(\left| q \right| =e=1.6\times { 10 }^{ -19 }C.\) The velocity acquired by the electron due to accelerating potential 200 V is
\({ v }_{ 200 }=\sqrt { \frac { 2\left( 1.6\times { 10 }^{ -19 } \right) \left( 200 \right) }{ \left( 9.1\times { 10 }^{ -31 } \right) } } =8.39\times { 10 }^{ 6 }m{ s }^{ -1 }\)
Since the speed v₂₀₀ > v, the electron is deflected towards direction of Lorentz force. So, in order to have null deflection, the potential, we have to supply is
\(v=\frac { { 1mv }^{ 2 } }{ 2\quad e } =\frac { \left( 9.1\times { 10 }^{ -31 } \right) \times \left( 7.23\times { 10 }^{ 6 } \right) ^{ 2 } }{ 2\times \left( 1.6\times { 10 }^{ -19 } \right) } \)
V = 148.65 V

The susceptibility of material X is
X_m,x = \(\frac { \left| \overset { \rightarrow }{ M } \right| }{ \left| \overset { \rightarrow }{ H } \right| } =\frac { 500 }{ 1000 } =0.5\)
The susceptibility of material Y is
X_{m,y =} \(\frac { \left| \overset { \rightarrow }{ M } \right| }{ \left| \overset { \rightarrow }{ H } \right| } =\frac { 2000 }{ 1000 } =2\)
Since, susceptibility of material Y is greater than that of material X, material Y can be easily magnetized than X.

The magnitude of the force between two poles is given by
\( F =k\frac { { q }_{ m_A }{ q }_{ { m }_{ B } } }{ { r }^{ 2 } } \)
(Given : F = 9 × 10^–3 N, r = 10 cm = 10 × 10^–2 m
Since q_mA = q_mB = q_m, we have
9 x 10^-3 = 10^-7 x \(\frac { { q }_{ m }^{ 2 } }{ { \left( 10\times { 10 }^{ -2 } \right) }^{ 2 } } \Rightarrow { q }_{ m }\) = 30NT^-1 

Let the length MN = y and the point P is on its perpendicular bisector. Let O be the point on the conductor as shown in figure. Therefore,
\(OM=ON=\frac { y }{ 2 } ,then\)
\(cos\varphi _{ 1 }=\frac { \frac { y }{ 2 } }{ \sqrt { \frac { { y }^{ 2 } }{ 4 } +{ d }^{ 2 } } } =\frac { adjacent \ length }{ hypotenuse \ length } \)
\(=\frac { ON }{ PH } =-\frac { \frac { y }{ 2 } }{ \sqrt { \frac { { y }^{ 2 } }{ 4 } +{ a }^{ 2 } } } =-\frac { y }{ \sqrt { { y }^{ 2 }+{ 4a }^{ 2 } } } \)
\(cos\varphi _{ 1 }=\frac { adjacent \ length }{ hypotenuse \ length } =\frac { OM }{ PM } \)
\(=-\frac { \frac { y }{ 2 } }{ \sqrt { \frac { { y }^{ 2 } }{ 4 } +{ a }^{ 2 } } } =-\frac { y }{ \sqrt { { y }^{ 2 }+{ 4a }^{ 2 } } } \)
Hence,
\(\overset { \rightarrow }{ B } =\frac { { \mu }_{ ° }I }{ 4\pi a\sqrt { { y }^{ 2 }+{ 4a }^{ 2 } } } \hat { n } \)
For long straight wire, Y\(\rightarrow \infty ,\)
\(\overset { \rightarrow }{ B } =\frac { { \mu }_{ ° }I }{ 2\pi a } \hat { n } \)
The result obtained is the same as we obtained in equation (3.39).

In a right angle triangle OPN let the angle \(\angle\)OPN = \(\theta \)₁ which implies, \({ \varphi }_{ 1 }=\frac { \pi }{ 2 } -{ \theta }_{ 1 }\) and also in a right angle triangle OPM, 
\(\angle\)OPN = \(\theta \)₂ which implies, \({ \varphi }_{ 2 }=\frac { \pi }{ 2 } +{ \theta }_{ 2 }\)
Hence,
\(\overset { \rightarrow }{ B } =\frac { { \mu }_{ ° }I }{ 4\pi a } \left( cos\left( \frac { \pi }{ 2 } -{ \theta }_{ 1 } \right) -cos\left( \frac { \pi }{ 2 } +{ \theta }_{ 2 } \right) \right) \hat { n } \)
\(=\frac { { \mu }_{ ° }I }{ 4\pi a } (si{ n }_{ 1 }+{ sin }_{ 2 })\hat { n } \)

\(tan \ I=\frac{B_V}{B_H}=1\)
∴ I = 45^o

Magnetic moment,
μ = IA
Angular momentum,
L = Iω
Ratio \(\frac{p_m}{L}=\frac{(q/T)\pi r^2}{mr^2\omega}=\frac{q}{2m}\)

Lorentz force F = Bqv
Energy w = qV
Energy is equal to kinetic energy,
\(qV=\frac{1}{2}mv^2\)
\(v=\sqrt { \frac {2qV }{ m } } \)
\(\therefore Lorentz \ force \ F= Bq\times \sqrt \frac{2qV}{m}=\sqrt { \frac { 2{ B }^{ 2 }{ q }^{ 3 }V }{ m } } \)