(i) V = Ir (without voltmeter)R_v
\(V'=\frac { Ir }{ r+{ R }_{ v } } { R }_{ v }=\frac { Ir }{ 1+\frac { r }{ { R }_{ v } } } \)
V'
(ii) Percentage error =\(\left( \frac { V-V' }{ V } \right) \times \ 100\)
\(\left( \frac { r }{ r+{ R }_{ v } } \right) \times 100\)
(iii) \({ R }_{ v }\rightarrow \infty ,V' \ ={ I }_{ 2r }=V\)

\(\delta ={ 45 }^{ o },\theta ={ 60 }^{ o },\delta '=?\) 
\(tan\delta '=\frac { tan\delta }{ cos\theta } =\frac { { tan45 }^{ o } }{ { cos60 }^{ o } } =2\) 
\(\delta '={ tan }^{ -1 }(2)={ 63 }^{ o }{ 26 }^{ ' }\) 

\(Given \ v=50Hz, \ { E }_{ 0 }=200V\)
\(\therefore \ \omega =2\pi v=2\pi \times 50=100\pi\)
\(\therefore \ General \ equation \ for \ instantaneous \ e.m.f.,\)
\(E={ E }_{ 0 } \ sin\omega t=200 \ sin \ 100\pi t,\)

There is no contradiction. Field lines inside the bar go away from S towards N. The next flux of B over any surface fully enclosing N or S must be identically zero. 

V = V1 + V2
Q = C1V = 6 x 10^-6  x 2 = 12\(\mu \)C
As C₂ is in series same amount of charge will also flow through it now V₂ = Q/C₂ = (12 x 10^-6) /(12 x 10^-6) = 1 volt
Total Battery voltage, V = 2 + 1 = 3 Volt

Biot-Savarts law states that \(d \mathbf{B}=\frac{\mu_0}{4 \pi} \cdot \frac{\mathrm{I} d \mathbf{l} \times \hat{\mathbf{r}}}{|\mathbf{r}|^2}\)
Here, \(\Delta\)x = w cm
\(\because\) \(\Delta\)l = \(\Delta\)x \(\hat{\mathbf{i}}\)
\(\Rightarrow\)   I = 2 A, r = 1 m
\(\begin{aligned} \therefore \quad d B=\frac{\mu_0}{4 \pi} \cdot \frac{(2 w \hat{\mathbf{i}} \times \hat{\mathbf{j}})}{(1)^2} \\ \end{aligned}\)
\(\begin{aligned} \mathrm{I} d l=2 \times w \hat{\mathrm{i}} \\ \end{aligned}\)
\(\begin{aligned} & \because \quad \hat{\mathbf{r}}=\hat{\mathbf{j}} \Rightarrow|\mathbf{r}|=1 \mathrm{~m} \\ \end{aligned}\)
\(\begin{aligned} & \therefore \quad d \mathrm{~B}=\frac{\mu_0 w}{2 \pi} \hat{\mathbf{k}} \\ \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad|d \mathrm{~B}|=\frac{\mu_0 w}{2 \pi} \\ \end{aligned}\)
and direction along + Z-axis.

The principle of galvanometer is based on the interaction of current and magnetic field.

In a parallel combination of resistors, the effective area of a cross-section of the conductor increases. As ​\(R∝l,\)​ therefore, resistance decreases.

(i) Respecting law and obeying the rules enacted by the nation; self discipline and avoiding arguments
(ii) The change in the magnetic field in a metal detector due to the metals carried by the contents of checking which, induces the change in current in the coil of the metal detector. This detection of change sets up an alarm.

(a) X-rays are produced by bombarding a metal target by high energy electrons.
(b) To study the atomic structures, treatment for certain forms of cancer.
(c) Presence of mind. alertness, taking initiative, helpful, caring.

Here, \(n=10^9\) electrons/sec
Charge given/sec,
\(q=ne=10^9\times 1.6\times 10^{-19}C\)
Total charge Q = 1C
Time required = \({Q\over q}={1\over 1.6\times 10^{-10}}sec\)
\(=6.25\times10^9s={6.25\times 10^9\over 3600\times 24\times 365}year\)
= 198.18 year

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, Maximum magnetic field,
B₀ = 5 x 10^-9T;
charge on alpha particle, q = + 2e
= 2 x 1.6 x 10^-19 = 3.2 x 10^-19C, 
v = 3 x 10⁷ ms^-1 
Maximum electric field,
E₀ = cB₀ = (3 x 10⁸) x (5 x 10^-19) = 150 Vm^-1
Maximum force on alpha particle due to electric field = q x E₀ = (3.2 x 10^-19) x 150
= 4.80 x 10^-17 N
Force on alpha particle due to magnetic field = qvB₀ = (3.2 x 10^-19) x ( 3 x 10⁷) x (5 x 10^-9) 
= 4.80 x 10^-19 N

Here, \(r=6\times { 10 }^{ -2 }m;\)
\(A=\pi \times \left( 6\times { 10 }^{ -2 } \right) ^{ 2 }=36\pi \times { 10 }^{ -4 }{ m }^{ 2 }\)
\(\frac { dE }{ dt } ={ 10 }^{ 10 }V{ cm }^{ -1 }{ s }^{ -1 }={ 10 }^{ 12 }V{ m }^{ -1 }{ s }^{ -1 }\)
Displacement current,
\({ I }_{ D }={ \epsilon }_{ 0 }\frac { d{ \phi }_{ E } }{ dt } ={ \epsilon }_{ 0 }A\frac { dE }{ dt } \)
\(=\left( 8.85\times { 10 }^{ -12 } \right) \times \left( 36\pi \times { 10 }^{ -4 } \right) \times { 10 }^{ 12 }\)
= 0.1 A