A simple pendulum made of mass 10 g and a metallic wire of length 10 cm is suspended vertically in a uniform magnetic field of 2 T . The magnetic field direction is perpendicular to the plane of oscillations of the pendulum. If the pendulum is released from an angle of $60^{\circ}$ with vertical, then maximum induced EMF between the point of suspension and point of oscillation is
$\_\_\_\_$ mV . (Take $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ )
Explanation:
The mass $(\mathrm{m})$ of the bob is $10 \mathrm{~g}=0.01 \mathrm{~kg}$
The length of wire is $\mathrm{l}=10 \mathrm{~cm}=0.1 \mathrm{~m}$.
The magnetic field is uniform in the region with strength, $B=2 T$.
The pendulum is released from angle $\theta=60^{\circ}$
The conducting rod of length 1 rotating around a fixed pivot point with a constant angular velocity $\omega$. The rotation takes place in a uniform magnetic field $B$ that is perpendicular to the plane of rotation.
Imagine a small element of length $d r$ on the rod located at a distance $r$ from the pivot point (axis of rotation).
The linear velocity v of this small element is related to the angular velocity $\omega$ by the relation:
$ \mathrm{v}=\mathrm{r} \omega $
The induced EMF ( $\mathrm{d} \varepsilon$ ) in a small conductor of length dr moving with velocity v perpendicular to a magnetic field $B$ is given by:
$ \mathrm{d} \varepsilon=\mathrm{B} \cdot \mathrm{v} \cdot \mathrm{dr} $
However, unlike a rod moving in a straight line, a rotating rod has different velocities at different points. The part near the pivot moves slowly, while the tip moves fastest as $\mathrm{v}=\mathrm{r} \omega \Rightarrow \mathrm{v} \propto \mathrm{r}$
To find the total EMF ( $\varepsilon$ ) across the entire rod, we integrate the small EMFs from the pivot ( $r=0$ ) to the tip of the $\operatorname{rod}(\mathrm{r}=\mathrm{l})$.
$ \varepsilon=\int_0^{\varepsilon} \mathrm{d} \varepsilon=\int_0^{\mathrm{r}} \mathrm{~B} \cdot \mathrm{v} \cdot \mathrm{dr}=\int_0^{\mathrm{r}} \mathrm{~B} \cdot(\omega \mathrm{r}) \cdot \mathrm{dr} $
$\Rightarrow $ $\varepsilon=\mathrm{B} \omega \int_0^1 \mathrm{rdr}$
$\Rightarrow $ $\varepsilon=\mathrm{B} \omega\left[\frac{\mathrm{r}^2}{2}\right]_0^1$
$\Rightarrow $ $\varepsilon=\mathrm{B} \omega\left(\frac{\mathrm{l}^2}{2}-\frac{0^2}{2}\right)$
$\Rightarrow $ $\varepsilon=\mathrm{B} \omega\left(\frac{\mathrm{l}^2}{2}\right)$
$\Rightarrow $ $ \mathrm{E}=\frac{1}{2} \mathrm{~B} \omega \mathrm{l}^2 $
To find the maximum induced EMF, we need to find the point where the angular velocity $\omega$ (and consequently the linear speed v ) is maximum. For a pendulum, the speed is highest at the lowest point of its swing (the equilibrium position).
If the pendulum is released from an angle $\theta$, then the lost in height is,
$\mathrm{h}=\mathrm{l}(1-\cos \theta)$
$\Rightarrow $ $\mathrm{h}=0.1 \times\left(1-\cos 60^{\circ}\right)$
$\Rightarrow $ $\mathrm{h}=0.1 \times(1-0.5)=0.05 \mathrm{~m}$
Using Conservation of Energy, when the pendulum is released from an angle $\theta$, it loses potential energy and gains kinetic energy as it swings down.
$ \mathrm{mg} \Delta \mathrm{~h}=\frac{1}{2} \mathrm{~m} \mathrm{v}_{\max }^2 $
$\Rightarrow $ $\mathrm{v}_{\max }=\sqrt{2 \mathrm{gh}}$
$\Rightarrow $ $\mathrm{v}_{\max }=\sqrt{2 \times 10 \times 0.05}$
$\Rightarrow $ $ \mathrm{v}_{\max }=\sqrt{1}=1 \mathrm{~m} / \mathrm{s} $
Using the relationship between linear velocity (v) and angular velocity ( $\omega$ ) is $\mathrm{v}=\omega \mathrm{l}$.
$ \omega_{\max }=\frac{v_{\max }}{l} $
$\Rightarrow $ $ \omega_{\max }=\frac{1}{0.1}=10 \mathrm{rad} / \mathrm{s} $
Substituting the values into the rotational EMF formula :
$ \mathrm{E}_{\max }=\frac{1}{2} \mathrm{~B} \omega_{\max } \mathrm{l}^2 $
$\Rightarrow $ $\mathrm{E}_{\max }=\frac{1}{2} \times 2 \times 10 \times(0.1)^2$
$\Rightarrow $ $\mathrm{E}_{\max }=1 \times 10 \times 0.01$
$\Rightarrow $ $\mathrm{E}_{\max }=0.1 \mathrm{~V}$
$\Rightarrow $ $\mathrm{E}_{\max }=0.1 \times 1000 \mathrm{mV}$
$\Rightarrow $ $ \mathrm{E}_{\max }=100 \mathrm{mV} $
Therefore, the maximum induced EMF is 100 mV .
Hence, the correct answer is $\mathbf{1 0 0}$.
A conducting circular loop is rotated about its diameter at a constant angular speed of $100 \mathrm{rad} / \mathrm{s}$ in a magnetic field of 0.5 T perpendicular to the axis of rotation. When the loop is rotated by $30^{\circ}$ from the horizontal position, the induced EMF is 15.4 mV . The radius of the loop is $\_\_\_\_$ mm.
$ \left(\text { Take } \pi=\frac{22}{7}\right) $
Explanation:
When a conducting coil rotates in a uniform magnetic field, the magnetic flux passing through the coil changes continuously with time. According to Faraday's Law of Electromagnetic Induction, this changing flux induces an electromotive force (EMF) in the loop.
As the coil rotates with a constant angular velocity $\omega$, the angle $\theta$ changes continuously with time $t$. If we assume the normal of the coil is parallel to the magnetic field at time $t=0$ (so $\theta=0$ ), the angle at any given time t is :
$ \theta=\omega t $
So, the magnetic flux through the conducting loop is,
$ \Phi_{\mathrm{B}}=\mathrm{B} \mathrm{~A} \cos (\theta)=\mathrm{B} \mathrm{~A} \cos (\omega \mathrm{t}) $
For a coil consisting of N tightly wound turns, the total magnetic flux linkage is simply N times the flux through a single loop :
$ \Phi_{\text {total }}=\mathrm{N} \mathrm{~B} \mathrm{~A} \cos (\omega \mathrm{t}) $
According to Faraday's Law of Electromagnetic Induction, the induced electromotive force (e) in a closed loop is equal to the negative rate of change of the total magnetic flux with respect to time :
$ \mathrm{e}=-\frac{\mathrm{d} \Phi_{\text {total }}}{\mathrm{dt}} $
$\Rightarrow $ $e=-\frac{d}{d t}[N B A \cos (\omega t)]$
$\Rightarrow $ $e=-N B A \frac{d}{d t}[\cos (\omega t)]$
$ e=N B A \omega \sin (\omega t) $
So, the formula for the instantaneous induced EMF (E) in a rotating coil is:
The number of turns, $\mathrm{N}=1$
Angular speed, $\omega=100 \mathrm{rad} / \mathrm{s}$.
Magnetic field strength, $\mathrm{B}=0.5 \mathrm{~T}$.
Instantaneous induced EMF when angle of rotation $\theta=30^{\circ}$ is $\mathrm{e}=15.4 \mathrm{mV}=15.4 \times 10^{-3} \mathrm{~V}$.
Let the radius of the circular loop is $r$, then the area (A) of a circular loop is $A=\pi r^2$
$\mathrm{e}=\mathrm{B}\left(\pi \mathrm{r}^2\right) \omega \sin (\theta)$
$\Rightarrow $ $\mathrm{r}^2=\frac{\mathrm{e}}{\mathrm{B} \pi \omega \sin (\theta)}$
$\Rightarrow $ $\mathrm{r}^2=\frac{15.4 \times 10^{-3}}{0.5 \times\left(\frac{22}{7}\right) \times 100 \times \sin \left(30^{\circ}\right)}$
$\Rightarrow $ $\mathrm{r}^2=\frac{15.4 \times 10^{-3} \times 7}{550}=1.96 \times 10^{-4} \mathrm{~m}^2$
$\Rightarrow $ $\mathrm{r}=\sqrt{1.96 \times 10^{-4}}=1.4 \times 10^{-2} \mathrm{~m}=14 \mathrm{~mm}$
Inductance of a coil with $10^4$ turns is 10 mH and it is connected to a dc source of 10 V with internal resistance of $10 \Omega$. The energy density in the inductor when the current reaches $\left(\frac{1}{e}\right)$ of its maximum value is $\alpha \pi \times \frac{1}{e^2} \mathrm{~J} / \mathrm{m}^3$. The value of $\alpha$ is
$\_\_\_\_$ .
$ \left(\mu_0=4 \pi \times 10^{-7} \mathrm{Tm} / \mathrm{A}\right) . $
Explanation:
When an inductor is connected to a DC voltage source (V) with a resistance (R), it initially opposes the flow of current. Over time, the current increases until it reaches a steady, maximum state.
Once the current is steady, the inductor acts like an ideal wire (zero resistance). Therefore, the maximum current ( $\mathrm{i}_{\text {max }}$ ) is determined by Ohm's Law :
$ \mathrm{i}_{\max }=\frac{\mathrm{V}}{\mathrm{R}} $
For given circuit :
Voltage $\mathrm{V}=10 \mathrm{~V}$
Resistance $\mathrm{R}=10 \Omega$
$ \mathrm{i}_{\max }=\frac{10}{10}=1 \mathrm{~A} $
The current in the circuit is $\frac{1}{\mathrm{e}}$ time the maximum value, $\mathrm{i}=\mathrm{i}_{\text {max }} \times\left(\frac{1}{\mathrm{e}}\right)$
$ \mathrm{i}=1 \times\left(\frac{1}{\mathrm{e}}\right)=\frac{1}{\mathrm{e}} \mathrm{~A} $
For a long solenoid, the energy stored in an inductor is: $\mathrm{U}=\frac{1}{2} \mathrm{Li}^2$
The inductance of a long solenoid is: $\mathrm{L}=\frac{\mu_0 \mathrm{~N}^2 \mathrm{~A}}{\mathrm{l}}$, where N is total turns, A is cross-sectional area, and l is length.
The volume of the cylindrical core of the solenoid is: $\mathrm{V}_{\mathrm{vol}}=\mathrm{A} \times 1$
Then the energy density formula is $\mathrm{u}=\frac{\operatorname{energy}(\mathrm{U})}{\operatorname{volume}\left(\mathrm{V}_{\text {vol }}\right)}$
$ \mathrm{u}=\frac{\frac{1}{2} \mathrm{Li}^2}{\mathrm{~A} \cdot \mathrm{l}} $
$\Rightarrow $ $u=\frac{\frac{1}{2}\left(\frac{\mu_0 N^2 A}{l}\right) i^2}{A \cdot l}$
$\Rightarrow $ $\mathrm{u}=\frac{\mu_0 \mathrm{~N}^2 \mathrm{i}^2}{2 \mathrm{l}^2}$
$\Rightarrow $ $ \mathrm{u}=\frac{1}{2} \mu_0 \mathrm{n}^2 \mathrm{i}^2 $
Where $\mathrm{n}=\frac{\mathrm{N}}{\mathrm{l}}$ is the number of turns per unit length.
Putting the given values into the derived energy density formula :
$ \begin{aligned} \mu_0=4 \pi \times 10^{-7} \mathrm{Tm} / \mathrm{A}, \mathrm{n}=10^4 \mathrm{~m}^{-1}, \mathrm{i}=\frac{1}{\mathrm{e}} \mathrm{~A} ;\end{aligned} $
$\Rightarrow $ $\mathrm{u}=\frac{1}{2} \times\left(4 \pi \times 10^{-7}\right) \times\left(10^4\right)^2 \times\left(\frac{1}{\mathrm{e}}\right)^2$
$\Rightarrow $ $\mathrm{u}=20 \pi \times 1 / \mathrm{e}^2 \mathrm{~J} / \mathrm{m}^3$
$\Rightarrow 20 \pi \times \frac{1}{\mathrm{e}^2}=\alpha \pi \times \frac{1}{\mathrm{e}^2}$
$ \Rightarrow \alpha=20 . $
Therefore, the value of $\alpha$ is 20.
Conductor wire ABCDE with each arm 10 cm in length is placed in magnetic field of $\frac{1}{\sqrt{2}}$ Tesla, perpendicular to its plane. When conductor is pulled towards right with constant velocity of $10 \mathrm{~cm} / \mathrm{s}$, induced emf between points A and E is ________ mV .
Explanation:
As field is uniform we can replace the bent wire with straight wire from A to B.
So EMF :
$\begin{aligned} & \varepsilon=\operatorname{Bv} \ell_{\mathrm{AB}} \\ & =\frac{1}{\sqrt{2}} \times \frac{10 \mathrm{~cm}}{5} \times 2\left(10 \sin 45^{\circ}\right) \mathrm{cm} \\ & \varepsilon=10 \mathrm{mV} \end{aligned}$
A conducting bar moves on two conducting rails as shown in the figure. A constant magnetic field B exists into the page. The bar starts to move from the vertex at time t = 0 with a constant velocity. If the induced EMF is E ∝ tn, then value of n is _________.
Explanation:
$\mathrm{E}=\ell \mathrm{vB}$
$E=\frac{2 x}{\sqrt{3}} \times v B$ and $x=v t$
$\mathrm{E}=\frac{2}{\sqrt{3}} \mathrm{v}^2 \mathrm{Bt}$
$E \propto t^1$
In the given circuit the sliding contact is pulled outwards such that electric current in the circuit changes at the rate of $8 \mathrm{~A} / \mathrm{s}$. At an instant when R is $12 \Omega$, the value of the current in the circuit will be ________ A.
Explanation:
To analyze the given circuit, we can use the equation for induced electromotive force (EMF) in an inductor. The formula is:
$ \varepsilon - \frac{L \, dI}{dt} - I \cdot R = 0 $
For this circuit, we have:
$ \varepsilon = 12 \, \text{V} $ (the applied EMF)
$ L = 3 \, \text{H} $ (the inductance of the inductor)
$ \frac{dI}{dt} = 8 \, \text{A/s} $ (the rate of change of current)
Assuming the direction of current is such that the change in current is negative, substitute these values into the equation:
$ 12 - 3 \times (-8) - I \times 12 = 0 $
Solving for $ I $ (the current in the circuit):
$ 12 + 24 - 12I = 0 $
$ 36 = 12I $
$ I = \frac{36}{12} = 3 \, \text{A} $
Thus, the current at the moment when the resistance $ R $ is $ 12 \, \Omega $ is $ 3 \, \text{A} $.
The current in an inductor is given by $\mathrm{I}=(3 \mathrm{t}+8)$ where $\mathrm{t}$ is in second. The magnitude of induced emf produced in the inductor is $12 \mathrm{~mV}$. The self-inductance of the inductor _________ $\mathrm{mH}$.
Explanation:
The induced emf ($\varepsilon$) in an inductor is given by Faraday's law of electromagnetic induction, which in its differential form for an inductor can be expressed as:
$\varepsilon = L \frac{dI}{dt}$
where:
- $\varepsilon$ is the induced emf in the inductor,
- $L$ is the inductance of the inductor,
- $\frac{dI}{dt}$ is the rate of change of current through the inductor.
Given that the current $I = (3t + 8)$, where $t$ is in seconds, we can find the rate of change of current by differentiating $I$ with respect to $t$.
$\frac{dI}{dt} = \frac{d}{dt}(3t + 8) = 3$
The given magnitude of induced emf is $12 \, \text{mV} = 12 \times 10^{-3} \, \text{V}$ (since $1\,\text{mV} = 10^{-3} \, \text{V}$).
Now, plug these values into the formula to find $L$:
$12 \times 10^{-3} = L \cdot 3$
Solving for $L$ gives:
$L = \frac{12 \times 10^{-3}}{3} = 4 \times 10^{-3} \, \text{H} = 4 \, \text{mH}$
Therefore, the self-inductance of the inductor is 4 mH.
Explanation:
Explanation:
$\mathrm{B}_0$ is the magnetic field at origin
$ \begin{aligned} & \frac{d B}{d x}=-\frac{10^{-3}}{10^{-2}} \\\\ & \int_{B_0}^B d B=-\int_0^x 10^{-1} d x \\\\ & B-B_0=-10^{-1} x \\\\ & B=\left(B_0-\frac{x}{10}\right) \end{aligned} $
Motional emf in $\mathrm{AB}=0$
Motional emf in $\mathrm{CD}=0$
Motional emf in $\mathrm{AD}=\varepsilon_1=\mathrm{B}_0 \ell \mathrm{v}$
Magnetic field on rod BC B
$ =\left(\mathrm{B}_0-\frac{\left(-12 \times 10^{-2}\right)}{10}\right) $
Motional emf in $\mathrm{BC}=\varepsilon_2=\left(\mathrm{B}_0+\frac{12 \times 10^{-2}}{10}\right) \ell \times \mathrm{v}$
$ \varepsilon_{\mathrm{eq}}=\varepsilon_2-\varepsilon_1=300 \times 10^{-7} \mathrm{~V} $
For time variation
$ \begin{aligned} & \left(\varepsilon_{\text {eq }}\right)^{\prime}=\mathrm{A} \frac{\mathrm{dB}}{\mathrm{dt}}=60 \times 10^{-7} \mathrm{~V} \\\\ & \left(\varepsilon_{\mathrm{eq}}\right)_{\text {net }}=\varepsilon_{\mathrm{eq}}+\left(\varepsilon_{\mathrm{eq}}\right)^{\prime}=360 \times 10^{-7} \mathrm{~V} \\\\ & \text { Power }=\frac{\left(\varepsilon_{\text {eq }}\right)_{\text {net }}^2}{\mathrm{R}}=216 \times 10^{-9} \mathrm{~W} \end{aligned} $
The magnetic flux $\phi$ (in weber) linked with a closed circuit of resistance $8 \Omega$ varies with time (in seconds) as $\phi=5 t^2-36 t+1$. The induced current in the circuit at $t=2 \mathrm{~s}$ is __________ A.
Explanation:
$\begin{aligned} & \varepsilon=-\left(\frac{\mathrm{d} \phi}{\mathrm{dt}}\right)=10 \mathrm{t}-36 \\ & \text { at } \mathrm{t}=2, \varepsilon=16 \mathrm{~V} \\ & \mathrm{i}=\frac{\varepsilon}{\mathrm{R}}=\frac{16}{8}=2 \mathrm{~A} \end{aligned}$
A small square loop of wire of side $l$ is placed inside a large square loop of wire of side $L\left(L=l^2\right)$. The loops are coplanar and their centers coincide. The value of the mutual inductance of the system is $\sqrt{x} \times 10^{-7} \mathrm{H}$, where $x=$ _________.
Explanation:
Flux linkage for inner loop.
$\begin{aligned} & \phi=\mathrm{B}_{\text {center }} \cdot \ell^2 \\ & =4 \times \frac{\mu_0 \mathrm{i}}{4 \pi \frac{\mathrm{L}}{2}}(\sin 45+\sin 45) \ell^2 \\ & \phi=2 \sqrt{2} \frac{\mu_0 \mathrm{i}}{\pi \mathrm{L}} \ell^2 \\ & \mathrm{M}=\frac{\phi}{\mathrm{i}}=\frac{2 \sqrt{2} \mu_0 \ell^2}{\pi \mathrm{L}}=2 \sqrt{2} \frac{\mu_0}{\pi} \\ & =2 \sqrt{2} \frac{4 \pi}{\pi} \times 10^{-7} \\ & =8 \sqrt{2} \times 10^{-7} \mathrm{H} \\ & =\sqrt{128} \times 10^{-7} \mathrm{H} \\ & \mathrm{x}=128 \end{aligned}$
A ceiling fan having 3 blades of length $80 \mathrm{~cm}$ each is rotating with an angular velocity of 1200 $\mathrm{rpm}$. The magnetic field of earth in that region is $0.5 \mathrm{G}$ and angle of dip is $30^{\circ}$. The emf induced across the blades is $\mathrm{N} \pi \times 10^{-5} \mathrm{~V}$. The value of $\mathrm{N}$ is _________.
Explanation:
$\begin{aligned} & B_v=B \sin 30=\frac{1}{4} \times 10^{-4} \\ & \omega=2 \pi \times f=\frac{2 \pi}{60} \times 1200 \mathrm{~rad} / \mathrm{s} \\ & \varepsilon=\frac{1}{2} B_V \omega \ell^2 \\ & =32 \pi \times 10^{-5} \mathrm{~V} \end{aligned}$
A horizontal straight wire $5 \mathrm{~m}$ long extending from east to west falling freely at right angle to horizontal component of earths magnetic field $0.60 \times 10^{-4} \mathrm{~Wbm}^{-2}$. The instantaneous value of emf induced in the wire when its velocity is $10 \mathrm{~ms}^{-1}$ is _________ $\times 10^{-3} \mathrm{~V}$.
Explanation:
$\begin{aligned} & \mathrm{B}_{\mathrm{H}}=0.60 \times 10^{-4} \mathrm{~Wb} / \mathrm{m}^2 \\ & \text { Induced emf e}=\mathrm{B}_{\mathrm{H}} \mathrm{v} \ell \\ &=0.60 \times 10^{-4} \times 10 \times 5 \\ &=3 \times 10^{-3} \mathrm{~V} \end{aligned}$
A square loop of side $10 \mathrm{~cm}$ and resistance $0.7 \Omega$ is placed vertically in east-west plane. A uniform magnetic field of $0.20 T$ is set up across the plane in north east direction. The magnetic field is decreased to zero in $1 \mathrm{~s}$ at a steady rate. Then, magnitude of induced emf is $\sqrt{x} \times 10^{-3} \mathrm{~V}$. The value of $x$ is __________.
Explanation:
$\begin{aligned} & \overrightarrow{\mathrm{A}}=(0.1)^2 \hat{\mathrm{j}} \\ & \overrightarrow{\mathrm{B}}=\frac{0.2}{\sqrt{2}} \hat{\mathrm{i}}+\frac{0.2}{\sqrt{2}} \hat{\mathrm{j}} \end{aligned}$
Magnitude of induced emf
$\mathrm{e}=\frac{\Delta \phi}{\Delta \mathrm{t}}=\frac{\overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{A}}-0}{1}=\sqrt{2} \times 10^{-3} \mathrm{~V}$
Two coils have mutual inductance $0.002 \mathrm{~H}$. The current changes in the first coil according to the relation $\mathrm{i}=\mathrm{i}_0 \sin \omega \mathrm{t}$, where $\mathrm{i}_0=5 \mathrm{~A}$ and $\omega=50 \pi$ rad/s. The maximum value of emf in the second coil is $\frac{\pi}{\alpha} \mathrm{~V}$. The value of $\alpha$ is _______.
Explanation:
$\begin{aligned} & \phi=\mathrm{Mi}=\mathrm{Mi}_0 \sin \omega \mathrm{t} \\ & \mathrm{EMF}=-\mathrm{M} \frac{\mathrm{di}}{\mathrm{dt}}=-0.002\left(\mathrm{i}_0 \omega \cos \omega \mathrm{t}\right) \\ & \mathrm{EMF}_{\max }=\mathrm{i}_0 \omega(0.002)=(5)(50 \pi)(0.002) \\ & \mathrm{EMF}_{\max }=\frac{\pi}{2} \mathrm{~V} \end{aligned}$
Take $\pi=\frac{22}{7}$
Explanation:
$\omega = 210 \cdot \frac{2\pi \mathrm{rad}}{60 \mathrm{s}} = 22 \mathrm{rad/s}$
Now, we can find the linear velocity $v$ of the tip of the rod:
$v = \omega r$
where $r$ is the length of the rod (0.2 m).
$v = 22 \mathrm{rad/s} \cdot 0.2 \mathrm{m} = 4.4 \mathrm{m/s}$
Now, we can find the emf developed between the center and the ring using the formula:
$\epsilon = \frac{1}{2} B\ell v$
where $B$ is the magnetic field (0.2 T), $\ell$ is the length of the rod (0.2 m), and $v$ is the linear velocity (4.4 m/s).
$\epsilon = \frac{1}{2} \cdot 0.2 \mathrm{T} \cdot 0.2 \mathrm{m} \cdot 4.4 \mathrm{m/s} = 0.088 \mathrm{V}$
To express this value in mV, we can simply multiply it by 1000:
$\epsilon = 0.088 \mathrm{V} \cdot 1000 = 88 \mathrm{mV}$
So the emf developed between the center and the ring is 88 mV.
An insulated copper wire of 100 turns is wrapped around a wooden cylindrical core of the cross-sectional area $24 \mathrm{~cm}^{2}$. The two ends of the wire are connected to a resistor. The total resistance in the circuit is $12 ~\Omega$. If an externally applied uniform magnetic field in the core along its axis changes from $1.5 \mathrm{~T}$ in one direction to $1.5 ~\mathrm{T}$ in the opposite direction, the charge flowing through a point in the circuit during the change of magnetic field will be ___________ $\mathrm{mC}$.
Explanation:
The magnetic flux through the circuit is proportional to the magnetic field through the core, so we can write $\phi=NBA$, where $N$ is the number of turns in the loop, $B$ is the magnetic field through the core, and $A$ is the cross-sectional area of the core.
As the magnetic field changes from $1.5\mathrm{~T}$ in one direction to $-1.5\mathrm{~T}$ in the opposite direction, the change in magnetic flux is $\Delta\phi=2NBA$.
The induced emf drives a current $I$ through the resistor in the circuit, and the current and the resistance are related by Ohm's law, which is $I=\mathcal{E}/R$. Substituting the expression for $\mathcal{E}$ into this equation, we get $I=-d\phi/dtR$.
The charge $Q$ that flows through the circuit during the change in magnetic field is given by $Q=\int Idt$. Substituting the expression for $I$ into this equation and integrating with respect to time, we get $Q=-\Delta\phi/R$, where $\Delta\phi$ is the change in magnetic flux and $R$ is the resistance of the circuit.
Substituting the given values into this expression, we get:
$Q=-\frac{2NBA}{R}=-\frac{2(100)(1.5)(24\times10^{-4})}{12}=-0.06\mathrm{~C}=-60\mathrm{~mC}$
Therefore, the charge flowing through a point in the circuit during the change of magnetic field is $60\mathrm{~mC}$, which is the same as the provided answer.
A conducting circular loop is placed in a uniform magnetic field of $0.4 \mathrm{~T}$ with its plane perpendicular to the field. Somehow, the radius of the loop starts expanding at a constant rate of $1 \mathrm{~mm} / \mathrm{s}$. The magnitude of induced emf in the loop at an instant when the radius of the loop is $2 \mathrm{~cm}$ will be ___________ $\mu \mathrm{V}$.
Explanation:
The problem involves a conducting circular loop placed in a uniform magnetic field with its plane perpendicular to the field. The radius of the loop is expanding at a constant rate, and we are asked to find the magnitude of the induced emf in the loop at an instant when the radius of the loop is $2 \mathrm{~cm}$.
The magnetic flux through a circular loop of radius $r$ and area $A = \pi r^2$ placed in a uniform magnetic field $B$ perpendicular to the plane of the loop is given by:
$\Phi_B = B A = B \pi r^2$
The induced emf in the loop is given by Faraday's law of electromagnetic induction:
$\mathcal{E} = -\frac{d\Phi_B}{dt}$
In this case, the radius of the loop is expanding at a constant rate of $10^{-3} \mathrm{~m/s}$, which means that the rate of change of the area of the loop is:
$\frac{dA}{dt} = \frac{d}{dt} (\pi r^2) = 2 \pi r \frac{dr}{dt} = 2 \pi (0.02 \mathrm{~m}) (10^{-3} \mathrm{~m/s}) = 4 \times 10^{-5} \mathrm{~m^2/s}$
The magnetic flux through the loop is changing at this rate, and the induced emf in the loop is given by:
$\mathcal{E} = \left|\frac{d\Phi_B}{dt}\right| = \left|\frac{dB}{dt} \frac{dA}{dt}\right| = \left|B \frac{dA}{dt}\right| = \left|0.4 \mathrm{~T} \times 4 \times 10^{-5} \mathrm{~m^2/s}\right| = 16 \pi \mu \mathrm{V}$
Therefore, the magnitude of the induced emf in the loop at an instant when the radius of the loop is $2 \mathrm{~cm}$ is $50.24 $ $ \simeq $ 50 $\mu \mathrm{V}$.
A metallic cube of side $15 \mathrm{~cm}$ moving along $y$-axis at a uniform velocity of $2 \mathrm{~ms}^{-1}$. In a region of uniform magnetic field of magnitude $0.5 \mathrm{~T}$ directed along $z$-axis. In equilibrium the potential difference between the faces of higher and lower potential developed because of the motion through the field will be _________ mV.
Explanation:
$ \begin{aligned} \Delta V & =(E . d) \\\\ & =(V B) \times 0.15 \\\\ & =2 \times \frac{1}{2} \times 0.15 \mathrm{~V} \\\\ & =0.15 \mathrm{~V}=150 \mathrm{mV} \end{aligned} $
The magnetic field B crossing normally a square metallic plate of area $4 \mathrm{~m}^{2}$ is changing with time as shown in figure. The magnitude of induced emf in the plate during $\mathrm{t}=2 s$ to $\mathrm{t}=4 s$, is __________ $\mathrm{mV}$.
Explanation:
$ \begin{aligned} & \varepsilon=\left|\frac{\mathrm{d} \phi}{\mathrm{dt}}\right|=\frac{\mathrm{d}(\mathrm{BA})}{\mathrm{dt}}=\frac{\mathrm{AdB}}{\mathrm{dt}} \\\\ & \varepsilon=\frac{4 \mathrm{~d}(2 \mathrm{t})}{\mathrm{dt}}=4 \times 2=8 \mathrm{mV} \end{aligned} $
A square loop of side $2.0 \mathrm{~cm}$ is placed inside a long solenoid that has 50 turns per centimetre and carries a sinusoidally varying current of amplitude $2.5 \mathrm{~A}$ and angular frequency $700 ~\mathrm{rad} ~\mathrm{s}^{-1}$. The central axes of the loop and solenoid coincide. The amplitude of the emf induced in the loop is $x \times 10^{-4} \mathrm{~V}$. The value of $x$ is __________.
$ \text { (Take, } \pi=\frac{22}{7} \text { ) } $
Explanation:
In this problem, a square loop is inside a long solenoid, and there's a varying current flowing through the solenoid. Because the current is changing, it induces a changing magnetic field inside the solenoid.
According to Faraday's law of electromagnetic induction, a changing magnetic field will induce an electromotive force (emf) in a loop placed in that field. In this case, the loop is the square loop inside the solenoid.
The formula used here is based on Faraday's law, which states that the induced emf in a loop is equal to the rate of change of magnetic flux through the loop. This is given by:
$ \text{emf} = -\frac{d \Phi}{dt} $
where $\Phi$ is the magnetic flux.
The magnetic field inside a solenoid is given by $B = \mu_0 n I$, where $\mu_0$ is the permeability of free space, $n$ is the number of turns per unit length in the solenoid, and $I$ is the current through the solenoid.
The magnetic flux through the square loop is then given by $\Phi = B \cdot A = \mu_0 n I A$, where $A$ is the area of the loop.
When the current is sinusoidal, i.e., $I(t) = I_0 \sin(\omega t)$, its derivative with respect to time is $dI/dt = I_0 \omega \cos(\omega t)$, where $\omega$ is the angular frequency.
Hence, the rate of change of flux becomes:
$ \frac{d \Phi}{dt} = \mu_0 n A \frac{dI}{dt} = \mu_0 n A I_0 \omega \cos(\omega t) $
The emf, which is equal to the negative of the rate of change of flux, will have a maximum value (the amplitude) when $\cos(\omega t) = 1$, giving:
$
\text{Emf amplitude} = \mu_0 n A I_0 \omega$
$ = 4\pi \times 10^{-7} \, \text{T m/A} \times \left(\frac{50}{10^{-2}}\right) \, \text{turns/m} \times (2 \times 10^{-2} \, \text{m})^2 \times 2.5 \, \text{A} \times 700 \, \text{rad/s}
$
which simplifies to:
$ \text{Emf amplitude} = 44 \times 10^{-4} \, \text{V} $
So, the value of $x$ in the question is $44$
A 1 m long metal rod XY completes the circuit as shown in figure. The plane of the circuit is perpendicular to the magnetic field of flux density 0.15 T. If the resistance of the circuit is 5$\Omega$, the force needed to move the rod in direction, as indicated, with a constant speed of 4 m/s will be ____________ 10$^{-3}$ N.
Explanation:
To move the rod with a constant velocity $v=4 \mathrm{~m} \mathrm{~s}^{-1}$
$\mathrm{F}_{\text {net }}$ on the rod should be zero.
$ \begin{aligned} F & =B i l=0.15\left(\frac{B l v}{R}\right) l \\\\ & =0.15\left(\frac{0.15 \times 1 \times 4}{5}\right) \times 1 \\\\ & =0.03 \times 0.15 \times 4 \\\\ & =180 \times 10^{-4} \mathrm{~N} \\\\ & =18 \times 10^{-3} \mathrm{~N} \end{aligned} $
Two concentric circular coils with radii $1 \mathrm{~cm}$ and $1000 \mathrm{~cm}$, and number of turns 10 and 200 respectively are placed coaxially with centers coinciding. The mutual inductance of this arrangement will be ___________ $\times 10^{-8} \mathrm{H}$. (Take, $\pi^{2}=10$ )
Explanation:
The magnetic field $B_2$ due to the current $I_2$ in the larger coil with 200 turns is given by:
$B_2 = \frac{N_2 \mu_0 I_2}{2r_2} = \frac{200 \mu_0 I_2}{2 \times 10}$
The magnetic flux $\phi_{1,2}$ through the smaller coil due to this magnetic field is given by:
$\phi_{1,2} = N_1 \vec{B}_2 \cdot \vec{A}_1 = N_1 N_2 \frac{\mu_0 I_2}{2 r_2} \cdot \pi r_1^2$
Since $\phi_{1,2} = MI_2$, we can solve for the mutual inductance $M$:
$M = \frac{N_1 N_2 \frac{\mu_0 I_2}{2 r_2} \cdot \pi r_1^2}{I_2}$
Substituting the given values for $r_1$, $N_1$, $r_2$, and $N_2$:
$M = \frac{10 \times 200 \times 4 \pi \times 10^{-7} \times \pi \times (0.01)^2}{2 \times 10}$
Simplifying the expression, we get:
$M = 4 \times 10^{-8} \mathrm{H}$
So, the mutual inductance between the two concentric coils is $4 \times 10^{-8} \mathrm{H}$.
As per the given figure, if $\frac{\mathrm{dI}}{\mathrm{dt}}=-1 \mathrm{~A} / s$ then the value of $\mathrm{V}_{\mathrm{AB}}$ at this instant will be ____________ $\mathrm{V}$.
Explanation:
From the circuit :
${V_A} - iR - {{Ldi} \over {dt}} - 12 = {V_B}$
$ \Rightarrow {V_A} - {V_B} = 2 \times 12 + 6( - 1) + 12$ volts
$ = 30$ volts
A certain elastic conducting material is stretched into a circular loop. It is placed with its plane perpendicular to a uniform magnetic field B = 0.8 T. When released the radius of the loop starts shrinking at a constant rate of 2 cms$^{-1}$. The induced emf in the loop at an instant when the radius of the loop is 10 cm will be __________ mV.
Explanation:
$=2 \mathrm{~B} \pi \mathrm{r} \frac{\mathrm{dr}}{\mathrm{dt}}=2 \times \pi \times 0.1 \times 0.8 \times 2 \times 10^{-2}$
$=2 \pi \times 1.6=\mathbf{1 0 . 0 6}~[$ rounding off $\mathbf{1 0 . 0 6}=\mathbf{1 0}]$
Three identical resistors with resistance R = 12$\Omega$ and two identical inductors with self inductance L = 5 mH are connected to an ideal battery with emf of 12 V as shown in figure. The current through the battery long after the switch has been closed will be _____________ A.
Explanation:
After long time, inductors are shorted.
Effective circuit becomes
Current through battery $=\frac{V}{\mathrm{R}_{\mathrm{eq}}}=\frac{12 \mathrm{~V}}{4 \Omega}=3 \mathrm{~A}$
where $\mathrm{R}_{\mathrm{eq}}=3$ resistors in parallel.
For the given circuit the current through battery of 6 V just after closing the switch 'S' will be _________ A.
Explanation:
Just after closing the switch, $i = {6 \over {4 + 2}} = 1\,A$
A conducting circular loop is placed in $X-Y$ plane in presence of magnetic field $\overrightarrow{\mathrm{B}}=\left(3 \mathrm{t}^{3} \,\hat{j}+3 \mathrm{t}^{2}\, \hat{k}\right)$ in SI unit. If the radius of the loop is $1 \mathrm{~m}$, the induced emf in the loop, at time, $\mathrm{t}=2 \mathrm{~s}$ is $\mathrm{n} \pi \,\mathrm{V}$. The value of $\mathrm{n}$ is ___________.
Explanation:
${B_ \bot } = 3{t^2}$
${{d{B_ \bot }} \over {dt}} = 6t - 12$ at $t = 2$
${{d{\phi _1}} \over {dt}} = 12 \times \pi {(1)^2} = 12\pi $
In a coil of resistance $8 \,\Omega$, the magnetic flux due to an external magnetic field varies with time as $\phi=\frac{2}{3}\left(9-t^{2}\right)$. The value of total heat produced in the coil, till the flux becomes zero, will be _____________ $J$.
Explanation:
$R = 8\,\Omega $
$\phi = {2 \over 3}(9 - {t^2})$
At $t = 3$, $\phi = 0$
$\varepsilon = \left| { - {{d\phi } \over {dt}}} \right| = {4 \over 3}t$
$H = \int_0^3 {{{{V^2}} \over R}dt = \int_0^3 {{1 \over 8} \times {{16} \over 9}{t^2}dt} } $
$ = {2 \over 9} \times \left( {{{{t^3}} \over 3}} \right)_0^3 = {2 \over {9 \times 3}} \times 27 = 2\,J$
Magnetic flux (in weber) in a closed circuit of resistance 20 $\Omega$ varies with time t(s) at $\phi$ = 8t2 $-$ 9t + 5. The magnitude of the induced current at t = 0.25 s will be ____________ mA.
Explanation:
$R = 20\,\Omega $
$\phi = 8{t^2} - 9t + 5$
$\varepsilon = \left| { - {{d\phi } \over {dt}}} \right| = |16t - 9| = |16(0.25) - 9| = 5$
$i = {\varepsilon \over R} = {5 \over {20}} = 0.25\,A = {{0.25} \over {{{10}^3}}} \times {10^3}\,A = 250\,mA$
A metallic rod of length 20 cm is placed in North-South direction and is moved at a constant speed of 20 m/s towards East. The horizontal component of the Earth's magnetic field at that place is 4 $\times$ 10$-$3 T and the angle of dip is 45$^\circ$. The emf induced in the rod is ___________ mV.
Explanation:
$E = Blv$
$ = 4 \times {10^{ - 3}} \times {{20} \over {100}} \times 20$ Volts
$ = 16$ mV
A 10 $\Omega$, 20 mH coil carrying constant current is connected to a battery of 20 V through a switch. Now after switch is opened current becomes zero in 100 $\mu$s. The average e.m.f. induced in the coil is ____________ V.
Explanation:
Initially current, ${I_0} = {{20} \over {10}} = 2A$ (when initially switch closed)
average emf induced in coil $ = {{Ldi} \over {dt}}$
$ \Rightarrow {e_{avg}} = {{20 \times {{10}^{ - 3}} \times (2 - 0)} \over {100 \times {{10}^{ - 6}}}}$
${e_{avg}} = 400\,V$
The current in a coil of self inductance 2.0 H is increasing according to I = 2 sin(t2) A. The amount of energy spent during the period when current changes from 0 to 2 A is ____________ J.
Explanation:
$U = {1 \over 2}L{I^2}$
$ = {1 \over 2}2 \times {2^2} = 4$ J
A circular coil of 1000 turns each with area 1m2 is rotated about its vertical diameter at the rate of one revolution per second in a uniform horizontal magnetic field of 0.07T. The maximum voltage generation will be ___________ V.
Explanation:
${V_{\max }} = NAB\omega $
$ = 1000 \times 1 \times 0.07 \times (2\pi \times 1)$
$ \simeq 440$ volts
[mp = 1.67 $\times$ 10$-$27 kg, e = 1.6 $\times$ 10$-$19C, Speed of light = 3 $\times$ 108 m/s]
Explanation:
Number of revolution = n
$n[2 \times {q_P} \times V] = {1 \over 2}{m_P} \times v_P^2$
$n[2 \times 1.6 \times {10^{ - 19}} \times 12 \times {10^3}]$
$ = {1 \over 2} \times 1.67 \times {10^{ - 27}} \times {\left[ {{{3 \times {{10}^8}} \over 6}} \right]^2}$
n(38.4 $\times$ 10$-$16) = 0.2087 $\times$ 10$-$11
n = 543.4
Explanation:
N = 20, $\omega$ = 50, B = 3 $\times$ 10$-$2 T
$\varepsilon $ = 20 $\times$ 50 $\times$ $\pi$ $\times$ (0.08)2 $\times$ 3 $\times$ 10$-$2 = 60.28 $\times$ 10$-$2
Rounded off to nearest integer = 60
The magnitude of current through R = 2$\Omega$ resistor at t = 5 s is ___________ mA.
Explanation:
$\left| i \right| = {{\left| \in \right|} \over R}$ = 10t + 10 mA
at t = 5
$\left| i \right|$ = 60 mA
$B = {4 \over \pi } \times {10^{ - 3}}T\left( {1 - {t \over {100}}} \right)$
The energy dissipated by the coil before the magnetic field is switched off completely is E = ___________ mJ.
Explanation:
$\phi = {4 \over \pi } \times {10^{ - 3}}\left( {1 - {t \over {100}}} \right).\pi {R^2}$
$\phi = 4 \times {10^{ - 3}} \times {(1)^2}\left( {1 - {t \over {100}}} \right)$
$\varepsilon = {{ - d\phi } \over {dt}}$
$\varepsilon = {{ - d} \over {dt}}\left( {4 \times {{10}^{ - 3}}\left( {1 - {t \over {100}}} \right)} \right)$
$\varepsilon = 4 \times {10^{ - 3}}\left( {{1 \over {100}}} \right) = 4 \times {10^{ - 5}}V$
When B = 0
$1 - {t \over {100}} = 0$
t = 100 sec
Heat $ = {{{\varepsilon ^2}} \over R}t$
Heat $ = {{{{(4 \times {{10}^{ - 5}})}^2}} \over {2 \times {{10}^{ - 6}}}} \times 100$ J
Heat $ = {{16 \times {{10}^{ - 10}} \times 100} \over {2 \times {{10}^{ - 6}}}}$ J
Heat = 0.08 J
Heat = 80 mJ
Explanation:
$ = 3t$
$L\int {di = 3\int {tdt} } $
$Li = {{3{t^2}} \over 2}$
$i = {{3{t^2}} \over {2L}}$
energy, $E = {1 \over 2}L{i^2}$
$ = {1 \over 2}L{\left( {{{3{t^2}} \over {2L}}} \right)^2}$
$ = {1 \over 2} \times {{9{t^4}} \over {4L}}$
$ = {9 \over 8} \times {{{{(4)}^4}} \over {4 \times 2}} = 144$ J
of 102 A s–1. The value of the potential difference VP – VQ , (in volts) at that instant, is _________.
Explanation:
VP + L.${{di} \over {dt}}$ - 30 + 2i = VQ
$ \Rightarrow $ VP + 50 $ \times $ 10-3(-102) - 30 + 2$ \times $1 = VQ
$ \Rightarrow $ VP - VQ = 35 - 2 = 33
placed in the XY plane. C1 has 500 turns, and
a radius of 1 cm. C2 has 200 turns and radius
of 20 cm. C2 carries a time dependent current
I(t) = (5t2 – 2t + 3) A where t is in s. The emf
induced in C1 (in mV), at the instant t = 1 s is
${4 \over x}$. The value of x is ___ .
Explanation:
Total flux $\phi = {N_1}{B_2}\pi {R_1}^2 = {N_1}{N_2}{{{\mu _0}I} \over {2{R_2}}}\pi {R_1}^2$
$ \therefore $ $\phi = {{500 \times 200 \times 4\pi \times {{10}^{ - 7}} \times (5{t^2} - 2t - 3)\pi {{({{10}^{ - 2}})}^2}} \over {2 \times 20 \times {{10}^{ - 2}}}}$
= ${{{{10}^5} \times 4{\pi ^2} \times {{10}^{ - 7}}(5{t^2} - 2t + 3) \times {{10}^{ - 4}}} \over {40 \times {{10}^{ - 2}}}}$
$ \Rightarrow $ $\phi = (5{t^2} - 2t + 3) \times {10^{ - 4}}$
We know, $e = \left| {{{d\phi } \over {dt}}} \right| = (10t - 2) \times {10^{ - 4}}$
At $t = 1\sec $
$e = 8 \times {10^{ - 4}} = 0.8\,mV = {{0.8} \over {10}} = {4 \over 5}$
$ \therefore $ $x = 5$
Explanation:
|emf| = $\left| {{{d\phi } \over {dt}}} \right|$ = BA$\omega$ sin $\omega t$
|emf|max = BA$\omega$ = BA${{2\pi } \over T}$
= ${{3 \times {{10}^{ - 5}} \times \pi \times {{\left( {0.1} \right)}^2} \times 2\pi } \over {0.4}}$
= 15 $ \times $ 10-6
= 15 $\mu $V
Explanation:
$ \Rightarrow $ L = ${V \over {\left| {{{di} \over {dt}}} \right|}}$ = ${{100} \over {{{0.25} \over {0.025 \times {{10}^{ - 3}}}}}}$ = 10 mH
Explanation:
$ \Rightarrow $ $\phi $ = (3 $ \times $ 25) + (4 $ \times $ 25) = 175 weber
Explanation:
$ |E|=\left|\frac{d \phi}{d t}\right| =\frac{d}{d t}\left(\left(B_0+\beta t\right) A\right) $
$ =\beta \times A $
$ =0.04 \mathrm{~V} $
So the circuit can be rearranged as
Using Kirchhoff's law we can write
$ \begin{aligned} & E=L \frac{d i}{d t}+\frac{q}{C} \\\\ & L \frac{d i}{d t}=E-\frac{q}{C} \\\\ & \text { Or } \frac{d^2 q}{d t^2}=-\frac{1}{L C}(q-C E) \end{aligned} $
Using SHM concept we can write
$ q=C E+A \sin (\omega t+\phi)\left(\text { where } \omega=\frac{1}{\sqrt{L C}}\right) $
at $t=0, q=0 \& i=0$
So $A=C E \& \phi=-\frac{\pi}{2}$
$ q=C E-C E \cos \omega t $
so $i=\frac{d q}{d t}=C E \omega \sin \omega t$
So,
$ i_{\max } =\frac{10^{-3} \times 0.04}{\sqrt{0.1 \times 10^{-3}}} $
$ =4 \mathrm{~mA} $
Explanation:
$\therefore$ $U = {1 \over 2}{L_1}I_1^2 + {1 \over 2}{L_2}I_2^2 + M{I_1}{I_2}$
$ \Rightarrow U = {1 \over 2} \times (10 \times {10^{ - 3}}){1^2} + {1 \over 2} \times (20 \times {10^{ - 3}}) \times {2^2} + (5 \times {10^{ - 3}}) \times 1 \times 2$
$ = 55$ mJ
[Assume the velocity of wire PQ remains constant (1 cm/s) after key S is closed. Given e-1 = 0.37, where e is base of the natural logarithm]
Explanation:
$e = (v \times B)dl = {10^{ - 2}} \times 1 \times {10^{ - 1}}$
e = ${10^{ - 3}}V$
${\tau _L} = LR = ({10^{ - 3}})(1) = {10^{ - 3}}s = 1\,ms$
$i = {i_0}(1 - {e^{ - t/{\tau _L}}}) = {{{{10}^{ - 3}}} \over 1}(1 - {e^{ - 1}})$
$i = {10^{ - 3}}(1 - 0.37)$
i = 0.63 mA
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 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:
$\begin{aligned} & \phi=n|\vec{B}||\vec{S}| \cos 45^{\circ}=2 \cdot \frac{\mu_0 i}{16 R} \cdot a^2 \cdot \frac{1}{\sqrt{2}}=\frac{\mu_0 i a^2}{2^{7 / 2} R}, \\\\ & M=\frac{\phi}{i}=\frac{\mu_0 a^2}{2^{7 / 2} R} .\end{aligned}$