\(\frac{\phi_1}{\phi_2} = \frac{2Q/\epsilon_0}{(2Q-Q)/\epsilon_0} \\ \Rightarrow 2 : 1 \)

The force acting on one wire due to currents through two wires is inversely proportional to the distance between them. Thus the force becomes 1/2 times if the distance between the wires is doubled.

When a cell is short-circuited, it gets totally discharged. The potential difference between the two plates of the cell becomes zero.

\(1.8\times 10^{-4}J\)

35.5\(\Omega\)
13.5V

The ratio of length to radius of the solenoid is quite large (about 35). Therefore, to estimate B approximately, we can use the exact result for a closely wound infinitely long solenoid.
(a) At the centre or near it,
 \(B={ \mu }_{ 0 }nI\) 
Where n is the number of turns per unit length.
Note 1. the radius of the wire does not enter this equation. Therefore, to get n, simply multiply number of turns per layer and divide the product by the length of the solenoid.
 \(n=\frac { 350\times 4 }{ 0.50 } =2800{ m }^{ -1 }\)
Now  \(I=6.0A,\)  
and  \({ \mu }_{ 0 }=4\pi \times { 10 }^{ -7 }T{ mA }^{ -1 },\)
Which gives \(B=2.1\times { 10 }^{ -2 }T\) 
Note 2. This estimate of B is for both on and off the axis, since for an infinitely long solenoid, the internal field near the centre is uniform over the entire cross-section.
(b) At the end of the solenoid,
\(B=\frac { { \mu }_{ 0 }nI }{ 2 }\)
\(=1.05\times { 10 }^{ -2 }T\)
(c) The outside field near the centre of long solenoid is negligible to the internal field.

(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.

In a cyclotron, when a proton crosses a region of potential difference V, the energy gained is eV. In cyclotron, in one revolution, the proton crosses the gap twice. So the energy gained by proton in each revolution = 2 eV.
Let the proton make n revolutions before emerging from the dees. The gain in its kinetic energy is 
\({ E }_{ k }=\frac { 1 }{ 2 } { mv }^{ 2 }=n\times 2eV \ or \ n=\frac { { mv }^{ 2 } }{ 4eV } =\frac { m{ \left( c/6 \right) }^{ 2 } }{ 4eV } \) 
\(or \ n=\frac { { mc }^{ 2 } }{ 4\times 36\times eV } \) 
\(=\frac { 1.67\times { 10 }^{ -27 }\times { \left( 3\times { 10 }^{ 8 } \right) }^{ 2 } }{ 4\times 36\times \left( 1.6\times { 10 }^{ -19 } \right) \times \left( 25\times { 10 }^{ 3 } \right) } \) 
= 261 revolutions. 

Here, force of attraction between charges = centripetal force 
\({1\over 4\pi\epsilon_o}{q_1q_2\over r^2}={mv^2\over r}\)
so \(v=\sqrt{{1\over 4\pi\epsilon_o}{q_1q_2\over mr}}\)
Time period of revolution
\(T={2\pi r\over v}=2\pi r\sqrt{4\pi\epsilon_om r\over q_1q_2}\)
\(T=\sqrt{16\pi ^3\epsilon_omr^3\over q_1q_2}\)