Electromagnetic Induction
Which of the following options is/are correct?
B = B0e${^{{{ - t} \over r}}}$ , where B0 and $\tau $ are constants, at time t = 0. If the resistance of the loop is R then the heat generated in the loop after a long time (t $ \to $ $\infty $) is :
Which of the following schematic plot(s) is (are) correct? (Ignore gravity)
Explanation:
The circuit is shown in the figure
Initially, right after the circuit is switched on ($ t \rightarrow 0^{+} $), the impedance (effective resistance) of inductors $ L_1 $ and $ L_2 $ is extremely high. Therefore, the inductors act as open circuits, and all the current flows through the resistor $ R = 12 \Omega $. According to Kirchhoff's law, the current through the battery at this moment is $ i_{\min} = \frac{V}{R} = \frac{5}{12} \, \text{A} $.
As the circuit reaches steady state ($ t \rightarrow \infty $), the impedance of the inductors drops to zero, and they function as resistors with their specified internal resistances.
The effective resistance of the circuit is $R_e=(12 \Omega \|$ $4 \Omega)\|3 \Omega=3 \Omega\| 3 \Omega=3 / 2 \Omega$.
In the steady state, the current through the circuit is $ i_{\max} = \frac{V}{R_e} = \frac{5}{\frac{3}{2}} = \frac{10}{3} \, \text{A} $.
Therefore, the ratio of the maximum current to the minimum current is:
$ \frac{i_{\max}}{i_{\min}} = \frac{\frac{10}{3}}{\frac{5}{12}} = \frac{10}{3} \times \frac{12}{5} = 8 $
A current carrying infinitely long wire is kept along the diameter of a circular wire loop, without touching it, the correct statement(s) is(are)
A circular wire loop of radius R is placed in the xy plane centred at the origin O. A square loop of side a(a << R) having two turns is placed with its centre at z = $\sqrt3$R along the axis of the circular wire loop, as shown in the figure. The plane of the square loop makes an angle of 45$^\circ$ with respect to z-axis. If the mutual inductance between the loops is given by ${{{\mu _0}{a^2}} \over {{2^{p/2}}R}}$, then the value of p is ___________.
Explanation:
The magnetic flux through a wire loop with $ n $ turns, an area vector $ \vec{S} $, placed in a uniform magnetic field $ \vec{B} $, is given by the equation $ \phi = n \vec{B} \cdot \vec{S} $. For a circular loop of radius $ r $ that carries a current $ i $, the magnetic field at an axial point located a distance $ z $ from the center of the loop can be expressed as :
$|\vec{B}|=\frac{\mu_0 i r^2}{2\left(r^2+z^2\right)^{3 / 2}}$
By substituting $ r = R $ and $ z = \sqrt{3} R $ into the expression for the magnetic field at an axial point of a circular loop, we get the magnetic field $ |\vec{B}| $ as follows:
The formula for the magnetic field at an axial point $ z $ from a circular loop of radius $ r $ carrying a current $ i $ is given by:
$ B = \frac{\mu_0 i r^2}{2 (r^2 + z^2)^{3/2}} $
Substituting $ r = R $ and $ z = \sqrt{3} R $ into this formula, we get:
$ B = \frac{\mu_0 i R^2}{2 (R^2 + (\sqrt{3} R)^2)^{3/2}} $
Simplify the expression inside the parentheses:
$ B = \frac{\mu_0 i R^2}{2 (R^2 + 3R^2)^{3/2}} $
$ B = \frac{\mu_0 i R^2}{2 (4R^2)^{3/2}} $
$ B = \frac{\mu_0 i R^2}{2 (4^{3/2} R^3)} $
$ 4^{3/2} = 8 $
$ B = \frac{\mu_0 i R^2}{2 (8 R^3)} $
$ B = \frac{\mu_0 i R^2}{16 R^3} $
$ B = \frac{\mu_0 i}{16 R} $
Thus, the magnetic field at the axial point $ z = \sqrt{3} R $ is:
$ |\vec{B}| = \frac{\mu_0 i}{16 R} $
This magnetic field is directed along the $ z $-axis and can be considered uniform at the location of the square loop (since $ a \ll R $). Consequently, $ \vec{B} $ forms an angle of $ 45^\circ $ with the area vector $ \vec{S} $ (where $ |\vec{S}| = a^2 $) of the square loop. The magnetic flux through the square loop and the mutual inductance of the loops are given by:
$\begin{aligned} & \phi=n|\vec{B}||\vec{S}| \cos 45^{\circ}=2 \cdot \frac{\mu_0 i}{16 R} \cdot a^2 \cdot \frac{1}{\sqrt{2}}=\frac{\mu_0 i a^2}{2^{7 / 2} R}, \\\\ & M=\frac{\phi}{i}=\frac{\mu_0 a^2}{2^{7 / 2} R} .\end{aligned}$
The figure shows certain wire segments joined together to form a coplanar loop. The loop is placed in a perpendicular magnetic field in the direction going into the plane of the figure. The magnitude of the field increases with time. $I_1$ and $I_2$ are the currents in the segments ab and cd. Then,
Two metallic rings A and B, identical in shape and size but having different resistivities $\rho_A$ and $\rho_B$, are kept on top of two identical solenoids as shown in the figure below. When current I is switched on in both the solenoids in identical manner, the rings A and B jump to heights $h_A$ and $h_B$, respectively, with $h_A > h_B$. The possible relation(s) between their resistivities and their masses $m_A$ and $m_B$ is (are)
$\left( {{\mu _0} = 4\pi \times {{10}^{ - 7}}\,Tm\,{A^{ - 1}}} \right)$
$\phi = 10{t^2} - 50t + 250$
The induced $emf$ at $t=3s$ is
What is the advantage of this system?
No friction hence no power consumption.
No electric power is used.
Gravitation force is zero.
Electrostatic force draws the train.
What is the disadvantage of this system?
Train experiences upward force according to Lenz's law
Friction force creates a drag on the train.
Retardation.
By Lenz's law, train experiences a drag.
Which force causes the train to elevate upwards
Electrostatic force.
Time-varying electric field.
Magnetic force.
Induced electric field
A long solenoid of radius a and number of turns per unit length $n$ is enclosed by cylindrical shell of radius R, thickness $d$ $(d < < R)$ and length L. A variable current $\mathrm{I}=\mathrm{I}_{0} \sin \omega t$ flows through the coil. If the resistivity of the material of cylindrical shell is $\mathrm{P}$, find the induced current in the shell.
