ii) We can destroy the magnetism of a magnet
a) by heating it
b) by applying magnetic field across it in reverse direction.

(ii) Here, M = 5A\({ m }^{ 2 }\)
2l = 0.20m and m = ?
As, M = m x 2l
\(\Rightarrow \)  m = M | 2l  = \(\frac { 5 }{ 0.2 } \) = 25A-m

Current sensitivity, \({ I }_{ s }=\frac { NAB }{ k } \) and
Voltage sensitivity, \(V_{ s }=\frac { NAB }{ kR } \)
Since, the resistance of the coil may vary, it implies an increase in current sensitivity may not necessarily increase voltage sensitivity.
Thus, the trajectory of both the particles will be same.

Here, I = 2.5 A, N = 100, l = 50 cm = 0.50 m
\(n=\frac{N}{l}=\frac{100}{0.50}=200\)
(i) B = \(\mu\)₀nI = 4\(\pi\)\(\times\)10^-7 \(\times\)200 \(\times\)2.5
B = 6.28 \(\times\) 10^-4 T
(ii) \(B=\frac{\mu_0 n I}{2}=\frac{4 \pi \times 10^{-7} \times 200 \times 2.5}{2}=3.14 \times 10^{-4} \mathrm{~T}\)

Here, v=2.18 x 10⁶ ms^-1,
r=5.3 x 10^-11 m, e=1.6 x 10^-19 C.
(a) Time period of revolution of electron is given by, 
\(T=\frac { 2\pi r }{ v } =\frac { 2\pi \times 5.3\times { 10 }^{ -11 } }{ 2.18\times { 10 }^{ 6 } } =1.528\times { 10 }^{ -16 }s\) 
Equivalent current, \(I=\frac { charge }{ time } =\frac { e }{ T } \) 
\(=\frac { 1.6\times { 10 }^{ -19 } }{ 1.528\times { 10 }^{ -16 } } =1.05\times { 10 }^{ -3 }A\) 
\((b)B=\frac { { \mu }_{ o } }{ 4\pi } \frac { 2\pi I }{ r } =\frac { { 10 }^{ -7 }\times 2\pi \times 1.05\times { 10 }^{ -3 } }{ 5.3\times { 10 }^{ -11 } } \) 
=12.4 T

Charge on alpha particle is +2 e. The revolving alpha particle is equivalent to current loop, having current
\(I=\frac { charge }{ time } =\frac { 2e }{ t } =\frac { 2\times 1.6\times { 10 }^{ -19 }C }{ 2\quad s } \) 
= 1.6 x 10^-19 A
Magnetic field at the centre of the circular loop is
\(B=\frac { { \mu }_{ o } }{ 4\pi } \frac { 2\pi I }{ r } =\frac { { \mu }_{ o }I }{ 2r } \) 
\(=\frac { \left( 4\pi \times { 10 }^{ -7 } \right) \times \left( 1.6\times { 10 }^{ -19 } \right) }{ 2\times 0.8 } \) 
\(=4\times \frac { 22 }{ 7 } \times \frac { { 10 }^{ -7 }\times 1.6\times { 10 }^{ -19 } }{ 2\times 0.8 } \) 
= 12.57x s10^-26 T

Let \({ r }_{ 0 }=\) Maximum radius of circular path followed by charged particle (Equal to the radius of the Dees)
 \({ v }_{ 0 }=\) Maximum velocity
Since the necessary centripetal force is provided by the Lorentz magnetic force, therefore,
\( \frac { { m{ v }_{ 0 } }^{ 2 } }{ { r }_{ 0 } } =Bq{ v }_{ 0 }\) 
\({ v }_{ 0 }=\frac { Bqr_{ 0 } }{ m }\)
\( \\ \therefore \ { K.E }_{ maxi }=\frac { 1 }{ 2 } \times m\times { \left( \frac { Bqr_{ 0 } }{ m } \right) }^{ 2 }\)
\(=\frac { { b }^{ 2 }{ q }^{ 2 }{ r_{ 0 } }^{ 2 } }{ 2m } \)
This is the required result.

A galvanometer is said to be sensitive, if it gives a large deflection, even when a small voltage is applied cross its coil.
Current sensitivity. It is defined as the deflection produced in the galvanometer on passing unit current through its coil. Therefore,
Current sensitivity \(=\frac { \theta }{ 1 } =\frac { nBA }{ k } \) 
Voltage sensitivity. It is defined as the deflection produced in produced in the galvanometer when a unit voltage is applied across its coil. Therefore V, then Voltage sensitivity \(=\frac { \theta }{ V } \) 
If R is resistance of coil and I is current that passes through coil on applying voltage V, then \(V=IR\)
 \(\therefore \) Voltage sensitivity \(=\frac { \theta }{ IR } =\frac { nBA }{ kR } \) 
Thus, a galvanometer will be highly sensitive, if (i) n is large ; (ii) B is large ; (iii) A is large ; (iv) R is small and (v) k is small.
However, n and A cannot be increased beyond certain limit otherwise, the sixe of the galvanomert and the resistance of the instrument will become large. Therefore, B is made as large as possible. To increase B, very strong permanent magnet is used. The suspension wire is made of phosphor bronze, as for this material, k is very small. The value of k further decreases, if the wire is hammered into flat strip. In very sensitive galvanometers, quartz k, is still smaller.

Consider two charges  \({ q }_{ 1 }\) and \({ q }_{ 2 }\) placed at a distance \(\left| { r }_{ 2 } \right| \) apart in air. The force between two charges
\(\left| \vec { { F }_{ e } } \right| =\frac { { q }_{ 1 }{ q }_{ 2 } }{ { 4\pi \varepsilon }_{ 0 } } \frac { 1 }{ { \left| { r }_{ 12 } \right| }^{ 2 } } \)  
Again consider two electrically neutral parallel current carrying elements of length  \({ dl }_{ 1 }\) and \({ dl }_{ 2 }\) carrying currents \({ I }_{ 1 }\) and \({ I }_{ 2 }\).
 \(\therefore \) Magnetic force between two current elements
\(\left| \vec { { F }_{ m } } \right| =\frac { { \mu }_{ 0 } }{ 4\pi } \frac { { I }_{ 1 }{ I }_{ 2 } }{ { \left| { r }_{ 12 } \right| }^{ 2 } } { dl }_{ 1 }{ dl }_{ 2 }\)
\( { I }_{ 1 }{ dl }_{ 1 }=\frac { { q }_{ 1 } }{ t } \times { dl }_{ 1 }={ q }_{ 1 }{ v }_{ 1 }\)
\({ I }_{ 2 }{ dl }_{ 2 }=\frac { { q }_{ 2 } }{ t } { dl }_{ 2 }={ q }_{ 2 }{ v }_{ 2 }\)
\( \left| \vec { { F }_{ m } } \right| =\frac { { \mu }_{ 0 } }{ 4\pi } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { \left| { r }_{ 12 } \right| }^{ 2 } } { v }_{ 1 }{ v }_{ 2 }\quad ...(2)\)
\(\frac { \left| \vec { { F }_{ m } } \right| }{ \left| \vec { { F }_{ e } } \right| } ={ v }_{ 1 }{ v }_{ 2 }.{ \mu }_{ 0 }{ \varepsilon }_{ 1 }...(3)\)  
Since L.H.S. is a dimensionless quantity therefore, the quantity \({ \mu }_{ 0 }{ \varepsilon }_{ 1 }\) must have dimensions of \({ \left( velocity \right) }^{ 2 }\) as numerator has dimensions of \({ \left( velocity \right) }^{ 2 }\) as \({ v }_{ 1 }\) and \({ v }_{ 2 }\) are the drift velocities of electrons in current elements \(\therefore{ v }_{ 1 }{ v }_{ 2 }={ 10 }^{ -5 }\times { 10 }^{ -5 }\) 
\(={ 10 }^{ -10 }{ m }^{ 2 }{ s }^{ -2 }\) 
Where \({ \mu }_{ 0 }{ \varepsilon }_{ 1 }={ 1/c }^{ 2 }\), where c is velocity of light \(\left( { c }^{ 2 }=9\times { 10 }^{ 16 }{ m }^{ 2 }{ s }^{ -2 } \right) \) 
From (3)

\(\therefore \frac { \left| \vec { { F }_{ m } } \right| }{ \left| \vec { { F }_{ e } } \right| } <1\)

 \(\left| \vec { { F }_{ m } } \right| <\left| \vec { { F }_{ e } } \right| \)