Here, n = 30, G = 100 \(\Omega \) , \(\varepsilon \) = 3 V,
R = 200, K = ?
Total resistance of the circuit
= G + R = 100 + 200 = 300 \(\Omega \) 
Current in the circuit which produces full scale deflection in the galvanometer,
\({ I }_{ g }=\frac { \varepsilon }{ G+R } =\frac { 3 }{ 300 } =\frac { 1 }{ 100 } A\) 
\(K=\frac { { I }_{ g } }{ n } =\frac { 1/100 }{ 30 } =\frac { 1 }{ 3 } \times { 10 }^{ -3 }A/division\) 
\(=\frac { 1 }{ 3 } \times { 10 }^{ -3 }\times { 10 }^{ 6 }\mu \quad A/division\) 
\(=333.3\mu \ A/division\) 

Given, I₁ = 4 A, I₂ = 6 A, r = 3 cm = 0.03 m
\(F=\frac{\mu_0}{4 \pi} \cdot \frac{2 I_1 I_2}{r}=\frac{10^{-7} \times 2 \times 4 \times 6}{0.03}\)
= 1.6 \(\times\)10^-4 N/m

(i) Magnetic field due to C at D is
\(B=\frac { { \mu }_{ o } }{ 4\pi } \frac { 2I }{ r } ={ 10 }^{ -7 }\times \frac { 2\times 3 }{ 0.10 } =6\times { 10 }^{ -6 }T\) 
(ii) Force on D, F = BI₁ lsin\(\theta \) = (6 x 10^-6) x 4 x (5 x 10^-2) x sin 90^o
 = 1.2 x 10^-6 N

Here, energy of electron,
E^' = 2000 eV = 2000 x 1.6 x 10^-19 J
= 3.2 x 10^-16 J
B = 0.2 T ; r = ?
As, \(E'=\frac { 1 }{ 2 } { mv }^{ 2 }\)   \(\therefore \ v=\sqrt { \frac { 2E' }{ m } } \) 
Also, \(Bev=\frac { m{ v }^{ 2 } }{ r } \quad \) 
or \(r=\frac { mv }{ Be } =\frac { m }{ Be } \sqrt { \frac { 2E' }{ m } } =\frac { \sqrt { 2E'm } }{ Be } \) 
\(r=\frac { \sqrt { 2\times 3.2\times { 10 }^{ -16 }\times 9\times { 10 }^{ -31 } } }{ 0.2\times 1.6\times { 10 }^{ -19 } } \) 
= 7.5 x 10^-4 m

Here, q = e = 1.6 x 10^-19 C, 
B = 2.5 T, v = 1.5 x 10⁷ ms^-1, \(\theta \) = 30^o 
F = qv B sin \(\theta \) = (1.6 x 10^-19) x (1.5 x 10⁷) x 25 x sin 30^o
= 3 x 10^-12 N.

Here, R = 0.8 cm = 8 x 10^-3 m ;
I = 100 A
\((a) \ { B }_{ inside }=\frac { { \mu }_{ o } }{ 4\pi } \frac { 2Ir }{ { R }^{ 2 } } \) 
\(={ 10 }^{ -7 }\times \frac { 2\times 100\times { \left( 0.2\times { 10 }^{ -2 } \right) }^{ 2 } }{ { \left( 8\times { 10 }^{ -3 } \right) }^{ 2 } } \) 
= 6.25 x 10^-4 T
\((b) \ { B }_{ surface }=\frac { { \mu }_{ o } }{ 4\pi } \frac { 2I }{ { R } } ={ 10 }^{ -7 }\times \frac { 2\times 100 }{ 8\times { 10 }^{ -3 } } \) 
= 2.5 x 10^-5 T
(c) When point is outside the wire,
r = 0.8 + 0.4 = 1.2 cm = 1.2 x 10^-2 m.
\({ B }_{ outside }=\frac { { \mu }_{ o } }{ 4\pi } \frac { 2I }{ { r } } ={ 10 }^{ -7 }\times \frac { 2\times 100 }{ 1.2\times { 10 }^{ -2 } } \) 
= 1.67 x 10^-5 T.

Here, dl= 1 cm = 10^-2 m,
r= (20 cm) = (0.20 m) and \(\overset { \rightarrow }{ r } =0.2\hat { i } \) 
\(I\overset { \rightarrow }{ dl } =(3\times { 10 }^{ -2 })\hat { j } \) 
\(\overset { \rightarrow }{ dB } =\frac { { \mu }_{ o } }{ 4\pi } \frac { I\overset { \rightarrow }{ dl } \times \overset { \rightarrow }{ r } }{ { r }^{ 3 } } \) 
\(=\frac { { 10 }^{ -7 }\times \left( 3\times { 10 }^{ -2 }\hat { j } \right) \times \left( 0.20\hat { i } \right) }{ { \left( 0.20 \right) }^{ 3 } } \) 
\(=7.5\times { 10 }^{ -8 }\left( \hat { j } \times \hat { i } \right) =7.5\times { 10 }^{ -8 }\left( -\hat { k } \right) \) 
\(=-\left( 7.5\times { 10 }^{ -8 }T \right) \hat { k } \) 
Thus magnitude of magnetic field is 7.5X10^-8 T and its direction is along negative Z-axis.

From Equation we find that there is an upward force F, of magnitude IlB,. For mid-air suspension, this must be balanced by the force due to gravity.
m g = I l B
\(B=\frac{m g}{I l}\)
\(=\frac{0.2 \times 9.8}{2 \times 1.5}=0.65 \mathrm{~T}\)
Note that it would have been sufficient to specify m / l, the mass per unit length of the wire. The earth’s magnetic field is approximately 4 x 10^–5 T and we have ignored it.

Given, total number of turns, N = 500
Length of solenoid, l = 0.5 m
Current, I = 5 A
Radius, r = 1 cm = 10^-2 m
Here, \(\begin{aligned} \frac{l}{r} & =\frac{0.5}{10^{-2}}=50 \Rightarrow l>>r \\ \end{aligned}\)
\(\begin{aligned} \therefore B & =\mu_0 n I=\frac{\mu_0 N I}{l} \\ \end{aligned}\)
\(\begin{aligned} =4 \pi \times 10^{-7} \times \frac{500}{0.5} \times 5 \end{aligned}\)
= 6.28 \(\times\) 10^-3 T