(ii) (a) Let the mass of proton = m; charge of proton = q, mass of a-particle = 4m
Charge of α-particle = 2q
Cyclotron frequency,
\(v=\frac{Bq}{2\pi m} \Rightarrow v \propto \frac{q}{m}\)
For proton frequency, v_p\(\propto \frac{q}{m}\)
For α-particle,
Frequency, v_a \(\propto \frac{2q}{4m}\)
or  v_a \(\propto \frac{q}{2m}\)
Thus, particles will not accelerate with same cyclotron frequency. The frequency of proton is twice than the frequency of α-particle.
(b) Velocity, \(v=\frac{Bqr}{m} \Rightarrow v \propto \frac{q}{m}\)
For proton velocity, \(v_p \propto \frac{q}{m}\)
For α-particle,
Velocity, v _a \(\propto \frac{2q}{4m}\) or v _a \(\propto \frac{q}{2m}\)
Thus, particles will not exit the dees with same velocity. The velocity of proton is twice than the velocity of α-particles.

Principle and working: A current carrying coil, placed in a uniform magnetic field, (can) experience a torque.
Consider a rectangular coil for which no. of turns = N
rea of cross-section = 1 x b = A,
Intensity of the uniform magnetic field = B,
Current through the coil = I
∴ Deflecting torque = \(BIl \times b=BIA\)
For N turns \(\tau\)  = NBIA
Restoring torque in the spring = k\(\theta\)
(k = restoring torque per unit twist)
\(\therefore NBIA=k\theta\)
\(\therefore I=(\frac{k}{NBA})\theta\)
\(\therefore I\alpha \theta\)
The deflection of the coil, is therefore, proportional to the current flowing through it.

(b) the soft iron core not only makes the field radial but also increase the strength of the magnetic field
(ii) We have
\(Current sensitivity =\frac{\theta}{I}=NBA/k\)
Voltage sensitivity =\(\frac{\theta}{V}=\frac{\theta}{IR}=(\frac{NBA}{K}).\frac{1}{R}{ 1/2}\)
It follows that an increase in current sensitivity may not necessarily increase the voltage sensitivity.

\(E=18keV=18\times { 10 }^{ 3 }\times 1.6\times { 10 }^{ -19 }J\)
[eV = 1.6 x10^-19J]
\( =8\times 1.6\times { 10 }^{ -16 }\)
Let velocity of an electron be v.
Magnetic field, B = 0.4 gauss =\(0.4\times { 10 }^{ -4 } \ T\)
Distance = 30 cm =0.3
Mass of electron, \({ m }_{ e }=9.1\times { 10 }^{ -31 }kg\)

Energy of electron, \(E=\frac { 1 }{ 2 } m{ v }^{ 2 }\)
\(18\times 1.6\times { 10 }^{ -16 }=\frac { 1 }{ 2 } \times 9.1\times { 10 }^{ -31 }{ v }^{ 2 }\)
\(v=0.795\times { 10 }^{ 8 }\quad m/s\)
Let the electron beam defects along a circular path of radius r.The magnetic force applied on the electron provides the centripetal force.
\(evB=\frac { { mv }^{ 2 } }{ r } \left[ { B }_{ y }F=q\left( v\times B \right) \right] \); between \(\left( v\times B \right) \) 90⁰
OA = r = OC;DC = d = 30cm
                    
\(r=\frac { mv }{ Be } =\frac { 9.1\times { 10 }^{ -31 }\times 0.795\times { 10 }^{ 8 } }{ 0.4\times { 10 }^{ -4 }\times 1.6\times { 10 }^{ -19 } } =11.3m\)
Let the deflection be AD = x.
From\(\Delta DCO, \ sin\theta =\frac { DC }{ OC } =\frac { d }{ r } =\frac { 0.3 }{ 11.3 }\Rightarrow \ \theta ={ 1.521 }^{ 0 }\)
and \(cos\theta =\frac { OD }{ OC } =\frac { OD }{ r } \quad or\quad OD=rcos\theta \)
Deflection at the end of the path,
\(AD=x=AO-OD=r-rcos\theta =r\left( 1-cos\theta \right) \)
or  x = 11.3(1-cos1.521⁰) = 0.0039m = 3.9mm
 x = 4 mm
Thus, the up or down deflection of the beam is approximately 4 mm.

(i) Figure shows, the longitudinal sectional View of long current carrying solenoid.

The current comes out of the plant of paper at point marked.B is the magnetic field at any point inside the solenoid.
Considering the rectangular closed path abcd.
applying Ampere's circuital law over loop abcd.
\(\oint { B } .dI={ \mu }_{ 0 }\times (total\ current\ passes\ through\ loop\ abcd)\)
\(\\ \int _{ a }^{ b }{ B } .D1+\int _{ b }^{ c }{ B } .d1+\int _{ c }^{ d }{ B } .d1+\int _{ d }^{ a }{ B } .d1={ \mu }_{ 0 }[(\frac { N }{ L } )li]\)
\(\\ where,\frac { N }{ L } =number\ of\ turns\ per\ unit\ length,\)
\(\\ ab=cd=l=length\ of\ rectangle\)
\(\\ \int _{ a }^{ b }{ Bdl } \cos { { 0 }^{ 0 } } +\int _{ b }^{ c }{ Bdl } Cos{ 90 }^{ 0 }+0+\int _{ b }^{ a }{ Bdl } cos{ 90 }^{ 0 }={ { \mu } }_{ 0 }(\frac { N }{ L } )li\)
\(\\ [\because cos{ 0 }^{ 0 }=1\quad and\quad cos{ 90 }^{ 0 }=0]\)
\(\\ B\int _{ a }^{ b }{ dl } ={ \mu }_{ 0 }(\frac { N }{ l } )li\)
\(\\ \Rightarrow B={ \mu }_{ 0 }(\frac { N }{ l } )i\quad or\quad B={ \mu }_{ 0 }ni\)
Where, a n = number of turns per unit length.
This is a required expression for magnetic field inside the long current carrying solenoid.
(ii)A solenoid bent into the form of closed loop is called toroid. The magnetic field B has a constant magnitude everywhere inside the toroid.

(a) Let magnetic field inside the toroid be B along the consider loop 1 as shown in the figure.
Applying ampere's circuital law,
\(\oint { B } .dI={ \mu }_{ 0 }(NI)\)
since, toroid of N turns, therads the loop 1, N times, each carrying current / inside the loop. Therefore, total current threading the loop.Therefore, total current threading the loop 1 is Ni
\(\Rightarrow { \oint { loop1 } }B.d1={ \mu }_{ 0 }\quad NI\Rightarrow B\oint { loop1 } d1={ \mu }_{ 0 }\quad NI\)
\(\\ B\times 2\pi r={ \mu }_{ 0 } \ NI \ or \ B=\frac { { \mu }_{ 0 }NI }{ 2\pi r } \)
(b)The magnetic field inside the open space interior of the toroid Let the loop 2 be which in the open space inside the toroid.
\( \therefore \) By Ampere's circuital law,
\(\oint { loop2 } B.dl={ \mu }_{ 0 }(\theta )=0\)
\(\\\Rightarrow d B=0\)
the magnetic field in the open space exterior of the toroid Let us consider a coplanar loop 3 in the open space of exterior of the toroid. Here, each turn of toroid threads the loop two times in opposite direction
therefore, net current threading the loop
\(=NI-NI=0\)
\(\\ \oint { loop3 } B.dl={ \mu }_{ 0 }(NI-NI)=0\)
\(\\ \Rightarrow B=0\)Thus, there is no magnetic field in the open space interior and exterior of the toroid.

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.

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.

On the basis of their in a magnetic field, the various materials can be classified in three classes.
(i) Diamagnetic: Those materials, which when placed in a magnetic field, are feebly magnetised in a direction opposite to the magnetising field are called diamagnetic substances.A few examples of diamagnetic materials are copper, zinc, bismuth, water, sodium chloride, helium, argon etc.
When a diamagnetic substance is suspended in a magnetic field, it arranges itself in the direction of the magnetic field.
(ii) Paramagnetic: Those materials, which when placed in a magnetic field, are feebly magnetised in the direction of magnetic field, are called paramagnetic substances.A few examples of paramagnetic substances are aluminium, sodium, antimony, platinum, copper chloride, liquid oxygen etc.
When a paramagnetic substance is suspended in a magnetic field it arranges itself to the direction of magnetic field.
(iii) Ferromagnetic: Those materials which when placed in a magnetic field are strongly magnetised in the direction of the magnetising field, are ferromagnetic substances.A few examples of ferromagnetic substances are iron, nickel, cobalt, alnico, mercury etc.

\(|\mathrm{dB}|=\frac{\mu_{0}}{4 \pi} \frac{I \mathrm{~d} l \sin \theta}{r^{2}}\)
dl = Δx = 10⁻²m , I = 10 A, r = 0.5 m = y, \(\mu_{0} / 4 \pi=10^{-7} \frac{\mathrm{T} \mathrm{m}}{\mathrm{A}}\) θ = 90° ; sin θ = 1
\(|\mathrm{dB}|=\frac{10^{-7} \times 10 \times 10^{-2}}{25 \times 10^{-2}}=4 \times 10^{-8} \mathrm{~T}\)
The direction of the field is in the +z-direction. This is so since
\(\mathrm{d} \mathbf{l} \times \mathbf{r}=\Delta x \hat{\mathbf{i}} \times y \hat{\mathbf{j}}=y \Delta x(\hat{\mathbf{i}} \times \hat{\mathbf{j}})=y \Delta x \hat{\mathbf{k}}\)
We remind you of the following cyclic property of cross-products
\(\hat{\mathbf{i}} \times \hat{\mathbf{j}}=\hat{\mathbf{k}} ; \hat{\mathbf{j}} \times \hat{\mathbf{k}}=\hat{\mathbf{i}} ; \hat{\mathbf{k}} \times \hat{\mathbf{i}}=\hat{\mathbf{j}}\)
Note that the field is small in magnitude.