Consider the given figure,

From the above figure, it can be seen that the direction of currents in a long straight conductor $$XY$$ and arm $$AB$$  of a square loop $$ABCD$$  are in the same direction. So, there exist a force of attraction between the two, which will be experienced by $${F_{BA}}$$ as
$${F_{BA}} = \frac{{{\mu _0}liL}}{{2\pi \left( {\frac{L}{2}} \right)}}$$
In the case of $$XY$$ and arm $$CD,$$  the direction of currents are in the opposite direction. So, there exist a force of repulsion which will be experienced by $$CD$$  as
$${F_{CD}} = \frac{{{\mu _0}liL}}{{2\pi \left( {\frac{{3L}}{2}} \right)}}$$
Therefore, net force on the loop $$ABCD$$   will be
$$\eqalign{ & {F_{{\text{loop}}}} = {F_{BA}} - {F_{CD}} = \frac{{{\mu _0}liL}}{{2\pi }}\left[ {\frac{1}{{\left( {\frac{L}{2}} \right)}} - \frac{1}{{\left( {\frac{{3L}}{2}} \right)}}} \right] \cr & {F_{{\text{loop}}}} = \frac{{2{\mu _0}iL}}{{3\pi }} \cr} $$

Net flux through solenoid is, $${\phi _{{\text{net}}}} = N\phi $$
$$\eqalign{ & \therefore {\phi _{{\text{net}}}} = 500 \times 4 \times {10^{ - 3}} \cr & = 2\,Wb \cr} $$
where, $$\phi =$$  flux through each turn, and
$$N =$$  total number of turns
Also, $${\phi _{{\text{net}}}} = Li = 2\,Wb$$
$$\eqalign{ & {\text{Now,}}\,\,L \times 2 = 2 \cr & \Rightarrow {\text{Self - inductance,}}\,L = 1\,H \cr} $$

$$\eqalign{ & {\text{Using}}\,{V_A} - {V_B} = RI + L\frac{{dI}}{{dt}} \cr & \Rightarrow 140 = 5R + 10L \cr & 60 = 5R - 10L \cr & \Rightarrow L = 4H \cr} $$

KEY CONCEPT : $$I = {I_0}\left( {1 - {e^{ - \frac{R}{L}t}}} \right)$$     (When current is in growth in $$LR$$  circuit)
$$ = \frac{E}{R}\left( {1 - {e^{ - \frac{R}{L}t}}} \right) = \frac{5}{5}\left( {1 - {e^{ - \frac{5}{{10}} \times 2}}} \right) = \left( {1 - {e^{ - 1}}} \right)$$

$$\eqalign{ & {\text{At}}\,x = 0,B = {B_0},{\text{so}}\,{e_1} = {B_0}{v_0}d \cr & {\text{At,}}\,x = d,B = {B_0}\left( {1 + \frac{d}{a}} \right),{\text{so}}\,{e_2} = {B_0}\left( {1 + \frac{d}{a}} \right){v_0}d \cr & {\text{Now}}\,{e_{{\text{net}}}} = {e_2} - {e_1} = \frac{{{B_0}{v_0}{d^2}}}{a} \cr} $$

Consider an elemental circle of thickness $$dr.$$
The induced emf in the circular path of radius $$r$$ is
$$\varepsilon = \frac{d}{{dt}}\left( {\pi {r^2}B} \right) = \pi {r^2}\alpha $$
The resistance of circular path is

The length of the path being $$2\pi r$$  and $$tdr$$  is the cross sectional area of current flow. For the element the power dissipated inside the path is
$$dP = \frac{{{\varepsilon ^2}}}{R} = \frac{{\pi t\sigma }}{2}{\alpha ^2}{r^3}dr$$
The total dissipated power $$P$$ is
$$P = \frac{{\pi t\sigma }}{2}{\alpha ^2}\int\limits_0^a {{r^3}dr} = \frac{{\pi t\sigma {a^4}}}{8}{\alpha ^2}$$

Total effective length crossing the flux lines $$ = \frac{{2vt}}{{\tan \phi }} + l$$
Current $$ = \frac{{B\left( {\frac{{2\,vt}}{{\tan \theta }} + l} \right)v}}{R}$$

The velocity component of conductor must be perpendicular to $${\vec \ell }$$ and $$\vec B.$$

The magnetic flux through the coil will be zero, and so induced emf in it will be zero.

$$\eqalign{ & e = - \frac{{d\phi }}{{dt}} = - \frac{{d\left( {N\overrightarrow B .\overrightarrow A } \right)}}{{dt}} \cr & = - N\frac{d}{{dt}}\left( {BA\cos \omega t} \right) = NBA\omega \sin \omega t \cr & \Rightarrow {e_{\max }} = NBA\omega \cr} $$