Tan \(\eth \) = V/H = \(\sqrt { 3 } \)
\(\eth \) = 60
As V = \(\surd 3H\) and
B² = V² + H² = 3H² + H² = 4H²
(0.4)² = 4H² therefore
H = 0.2 G

ML^-1T^-2

Work done in moving a unit positive charge along distance \(\delta l\)
\(\left| { E }_{ l } \right| \delta ={ V }_{ A }-{ V }_{ B\\ }\)
\(=V-(V+\delta V)\)
  \(=-\delta V\)
\(\left| { E }_{ l } \right| =-\frac { \delta V }{ \delta l } \)

Electric field is in the direction in which the potential decreases the steepest.
(ii) The magnitude of electric field is given by the change in the magnitude of potential per unit displacement, normal to the equipotential surface at the point.

Path d is followed by electric field lines. Electric field intensity inside the metallic sphere will be zero, therefore, no electric lines of force exist inside the sphere. Also, electric field lines are always perpendicular to the surface of the conductor.

emf induced
e = Blv
= 0.1 x 10 x 10^-2 x 20 V
= 0.2 volt
(ii) Current in the loop
\(i=\frac{e}{R} \) 
\(=\frac{0.2}{2}A=0.1A\)

\({ X }_{ c }=\frac { 1 }{ \omega C } =\frac { 1 }{ 2\pi vC } \)
AsC decreases, Xc will increase. Hence brightness will decrease.
(ii)
\({ X }_{ c }=\frac { 1 }{ \omega C } =\frac { 1 }{ 2\pi vC } \)
As frequence (v) decreases, Xc will increase. Hence brightness will decrease

Work done in bringing the charge q₁ from infinity to position r₁
W₁ = q₁ V(r₁)
Work done in bringing charge q₂ to the position r₂
W₂ = q₂ V(r₂) + \(\frac{q_1q_2}{4\pi\epsilon_0r_{12}}\)
Hence, total work done in assembling the two charges
W = W₁ +W₂
= q₁V(r₁) + q₂V(r₂) + \(\frac{q_1q_2}{4\pi\epsilon_0r_{12}}\)

Work done, W = q x \triangleV
But \( \triangle V = 0\) as the two diagonally opposite points are at the same potential due to 10 \muC charge.
\(\therefore W = 2 \mu C x 0 = 0\)    
Work done W = 0

Consider the figure, |P_A| = P_C = P

The magnitude of resultant P_R ,
\({ P }_{ R }=\sqrt { { { p }_{ 1 } }^{ 2 }+{ { p }_{ 2 } }^{ 2 }+2{ p }_{ 1 }{ p }_{ 2 }\cos { \theta } } \)
\(=\sqrt { { p }_{ 2 }+{ p }_{ 2 }+2{ p }^{ 2 }\cos { \theta } } \)
\(=\sqrt { { 2p }^{ 2 }(1+\cos { \theta } } \)
\(=\sqrt { 2{ p }^{ 2 }\times 2{ cos }^{ 2 }\frac { \theta }{ 2 } } =2p \ cos\frac { \theta }{ 2 } \)
\(\tan\alpha =\frac { { p }_{ 2 }sin\theta }{ { p }_{ 1 }+{ p }_{ 2 }cos\theta } =\frac { psin{ 120 }^{ 0 } }{ p+pcos{ 120 }^{ 0 } } \)
\(=\frac { p\sqrt { 3/2 } }{ p } =\sqrt { 3 }\)
\(p-\frac { p }{ 2 } \)
\(|{ P }_{ R }|=2pcos\frac { \theta }{ 2 } =2pcos\frac { { 120 }^{ 0 } }{ 2 } =2p\times \frac { 1 }{ 2 } =p\)
\({ P }_{ R } \ will \ subtend \ an \ angle \ of \ { 30 }^{ 0 } \ with \ X \ axis.\)
Now torque acting on the system.
\(\tau ={ P }_{ R }\times E={ p }_{ R }Esin\theta =\frac { 1 }{ 2 } { p }E\)
The torque will work to align the dipole in the direction of electric field E.

(i) The direction of electric field is perpendicular to their equipotential surface. So,  the direction of electric field is along X-axis as its length should be perpendicular to equipotential surface lying in YZ- plane.
(ii) Length of the dipole - 2b
As dipole's axis is along the Y-axis.
\(\therefore \)  Electric dipole moment,
p = q (2b)\(\overset { \wedge }{ j } \)
Electric field, E = E\(\overset { \wedge }{ i } \)
\(\because \)   \(\tau \) = p x E
= q(2b)\(\overset { \wedge }{ j } \) x E\(\overset { \wedge }{ i } \)
= +2 qbE \((\overset { \wedge }{ j } \times\overset { \wedge }{ i } )\)
= 2 qbE (-\(\overset { \wedge }{ k } \))
\(\therefore \) Torque, |\(\tau \)| = 2 qbE.

Initially,
\(T=2\pi \sqrt { \frac { I }{ 3{ B }^{ ' } } } and \ finally, \ { T }^{ ' }=2\pi \sqrt { \frac { I }{ m(B+{ B }^{ ' } } } \)
where, B^' = horizontal magnetic field = \(24\mu \)T
and     B  = magnetic field due to downward conductor
 \(=\frac { { \mu }_{ 0 } }{ 4\pi } .\frac { 2i }{ a } \quad =18\mu \)T
\(\therefore \frac { { T }^{ ' } }{ T } =\sqrt { \frac { { B }^{ ' } }{ B+{ B }^{ ' } } } \Rightarrow \frac { { T }^{ ' } }{ 0.1 } =\sqrt { \frac { 24 }{ 18+24 } } \Rightarrow { T }^{ ' }=0.076\)s

Potential is maximum at A as potential decreases in the direction of field or we can say that V_A > V_B = V_C

We can see that magnetic field is perpendicular to paper and current in the loop is in clockwise direction. So, by Fleming's left hand rule, force on each element of the loop is radially outwards, so loop will have a tendency to expand.

If any arbitrary surface encloses a dipole, then the net charge is zero because the total charge on the dipole is zero (dipole consists oftwo equal and opposite charges). According to Gauss' law,
\(\text { total flux }=\frac{1}{\varepsilon_0} \times \text { charge enclosed }=\frac{1}{\varepsilon_0} \times(0)=0 \Rightarrow \phi=0\).

(i) Capacitors are used in radio circuits for tuning purposes.
(ii) Capacitors are used in power supplies for smoothing the rectified current.

Magnetic lines of force are closed, continuous curves, but electric lines of force are continuous loop which are not closed.

Bt the method of production and the energy they posses.

Here, A =  \(20cm\times 30cm\)
= \(6\times { 10 }^{ -2 }{ m }^{ -2 }\)
B = 0.3T
Let \(\theta \) be the angle made by field B with the normal to the plane of the coil.
(i) Here, \(\theta \) = 90\(^{0}\)- 90\(^{0}\) = 0\(^{0}\)
So, flux, \(\phi \)= BA cos\(\theta \)
\(\phi =0.3\times 6\times { 10 }^{ -2 }\times cos{ 0 }^{ \circ }\)
\(=1.8\times { 10 }^{ -2 }Wb\)  (1)
(ii) Here, \(\theta \) = 90\(^{0}\)- 30\(^{0}\)
= 60\(^{0}\)
\(\phi =0.3\times 6\times { 10 }^{ -2 }\times cos{ 60 }^{ \circ }\\ \phi =0.9\times { 10 }^{ -2 }Wb\)
(iii) Here, \(\theta \) = 90\(^{0}\)
\(\phi =0.3\times 6\times { 10 }^{ -2 }\times cos{ 90 }^{ \circ }\\ \phi =0\)

(i) The total charge of the capacitor remains conserved on introduction of dielectric slab. Also, the capacitance of capacitor increases to K times of original values. 
\(\therefore \\ \)  CV = C' V '= (KC) V' \(\Rightarrow V'=\frac { V }{ K } \) 
\(\therefore \\ \) New electric field,
\(E'=\frac { V' }{ d } =\left( { \frac { { V }/{ K } }{ d } } \right) =\left( \frac { V }{ d } \right) \frac { 1 }{ k } =\frac { E }{ K } \) 
​​\(\therefore \\ \)  On introduction of dielectric medium, new electric field E' becomes \(\frac { 1 }{ K } \) times of its original value (decrease).
(ii) The thickness of dieletric slab is \(\frac{d}{2}, \text { i.e. } t=\frac{d}{2}\)
The capacitance of a capacitor due to dielectric slab,
\(\begin{aligned} C & =\frac{\varepsilon_0 A}{d-t+\frac{t}{K}} \end{aligned}\)
\(\begin{aligned} =\frac{\varepsilon_0 A}{d-\frac{d}{2}+\frac{d}{2 K}}=\frac{2 \varepsilon_0 A}{d\left(1+\frac{1}{K}\right)} \end{aligned}\)

Given, input voltage, V_p = 2.5 x 10³ V​​​
Input current, I_P = 20 A
Also, \(\frac { { N }_{ p } }{ { N }_{ s } } =\frac { 10 }{ 1 } \Rightarrow \frac { { N }_{ s } }{ { N }_{ p } } =\frac { 1 }{ 10 } \)​​
Percentage efficiency \(=\frac { Output\ power }{ Input\ power } \times 100\)
\(\Rightarrow \frac { 90 }{ 100 } =\frac { Output\ power }{ { V }_{ p }{ I }_{ p } } \)
(i) Output power \(=\frac { 90 }{ 100 } \times ({ V }_{ p }{ I }_{ p })\)
\(=\frac { 90 }{ 100 } \times(2.5\times{ 10 }^{ 3 }V)\times(20A)\)
\(=4.5\times{ 10 }^{ 4 }W\)
(ii) \(\therefore \ \frac { { V }_{ s } }{ { V }_{ p } } =\frac { { N }_{ s } }{ { N }_{ p } } \Rightarrow { V }_{ s }=\frac { { N }_{ s } }{ { N }_{ p } } \times { V }_{ p }\)
\(Voltage,\ { V }_{ s }=\frac { 1 }{ 10 }\times 2.5\times{ 10 }^{ 3 }V=250V\)
(iii) \({ V }_{ s }{ I }_{ s }=4.5\times { 10 }^{ 4 } \ W\)
\(Current,{ I }_{ s }=\frac { 4.5\times{ 10 }^{ 4 } }{ { V }_{ s } } =\frac { 4.5\times10^{ 4 } }{ 250 } =180 \ A\)

Here, \(R=40\Omega ,\ L=3\ mH=3\times 10^{ -3 }H\)
\(\ C=2\ \mu F=2\times 10^{ -6 }F,\ E_{ v }=110V,\ v=5000\ Hz\)
\(X_{ L }=\omega L=2\pi vL=2\times 3.14\times 5000\times 3\times 10^{ -3 }\)
 \(=94.2\Omega \)
\(X_{ C }=\frac { 1 }{ \omega C } =\frac { 1 }{ 2\pi vC }\)
\(=\frac { 1 }{ 2\times 3.14\times 5000\times 2\times 10^{ -6 } } =15.92\Omega\)
\(Z=\sqrt { R^{ 2 }+(X_{ L }-X_{ C })^{ 2 } }\)
\(=\sqrt { 40^{ 2 }+(94.2-15.92)^{ 2 } } =87.9\Omega \)
\(I_{ v }=\frac { E_{ v } }{ Z } =\frac { 110 }{ 87.9 } =1.25 \ A\)

\(8 \times { 10 }^{ -5 }Wb;8 \times { 10 }^{ -3 }Wb\)
\(Here,n=100,L=20mH=20 \times { 10 }^{ -3 }H;\)
\( i=4 \ mA=4 \times { 10 }^{ -3 }A\)
Magnetic flux through the cross section of coil 
\(Here,n=100,L=20mH=20 \times { 10 }^{ -3 }H;\)
\(Li=20 \times { 10 }^{ -3 } \times 4 \times { 10 }^{ -3 }=8 \times { 10 }^{ -5 }Wb\)
\(Total \ magnetic \ flu \times =nLi=100 \times 8 \times { 10 }^{ -5 }\)
\(=8 \times { 10 }^{ -3 }Wb\)

Given, E = 50 x \(\widehat { i } \)
and  \(\triangle\)S = 25 cm²
=  25 \(\times\) 10^-4 m²
As the electric field is only along the X-axis, so flux will pass only through the cross-section of cylinder.
Magnitude of electric field at cross-section A,
E_A = 50 \(\times\) 1 = 50 N C^-1
Magnitude of electric field at cross-section B,
E_B = 50 \(\times\) 2 = 100 NC^-1             
The corresponding electric fluxes are
\({ \phi }_{ A }\) = E _A . \(\triangle\)S = 50 \(\times\) 25 \(\times\) 10^-4 \(\times\) \(cos180^{o}\)
= -0.125 N-m² C^-1
\({ \phi }_{ B }\) = E _B .  \(\triangle\)S = 100 \(\times\) 25 \(\times\) 10^-4 \(\times\) \(cos0^{o}\)
= 0.25 N - m² C^-1
So, the net flux through the cylinder,
\(\phi\) = \({ \phi }_{ A }\) + \({ \phi }_{ B }\) = -0.125 + 0.25
= 0.125 N - m² C^-1    
(ii) Using Gauss' law,
\(\oint\) E. dS = \(\frac { q }{ { \varepsilon }_{ 0 } }\)   [\(\therefore\) \(\oint\)E . dS = \(\phi\)]
\(\Rightarrow\)  0.125 = \(\frac { q }{ { 8.85\times { 10 }^{ -12 } } } \)
\(\Rightarrow\) q = 8.85 \(\times\) 0.125 \(\times\) 10^-12 
= 1.1 \(\times\) 10^-12 C .