Explanation:
$\begin{aligned} & \mathrm{dM}=\mathrm{dIA} \\ & \mathrm{A}=\pi \mathrm{r}^2=\pi(\mathrm{R} \sin \theta)^2 \\ & \mathrm{dI}=\frac{\mathrm{da}}{\mathrm{T}}=\frac{\sigma(2 \pi \mathrm{r})(\mathrm{R} \mathrm{d} \theta) \omega}{2 \pi} \\ & \mathrm{dI}=\frac{\sigma 2 \pi \mathrm{R}^2 \omega \sin \theta \mathrm{~d} \theta}{2 \pi}\end{aligned}$
$ \mathrm{dI}=\sigma \mathrm{R}^2 \omega \sin \theta \mathrm{~d} \theta $
Magnetic dipole moment :
$ \begin{aligned} & M=\int d M=\int_0^\pi \sigma R^2 \omega \pi R^2 \sin ^3 \theta d \theta \\ & M=\sigma R^4 \omega \pi \int_0^\pi \sin ^3 \theta d \theta \quad\left(\because \int_0^\pi \sin ^3 \theta d \theta=\frac{4}{3}\right) \\ & M=\left(\frac{Q}{4 \pi R^2}\right) R^4 \omega \pi\left(\frac{4}{3}\right) \end{aligned} $
Magnetic dipole moment
$ \mathrm{M}=\frac{\mathrm{QR}^2 \omega}{3} $
Angular momentum
$ \begin{aligned} & \mathrm{L}=\left(\frac{2}{5} \mathrm{MR}^2\right) \omega \\ & \frac{\mathrm{M}}{\mathrm{~L}}=\frac{\mathrm{QR}^2 \omega}{3 \times \frac{2}{5} \mathrm{MR}^2 \omega}=\frac{\mathrm{Q}}{2 \mathrm{M}}\left(\frac{5}{3}\right) \end{aligned} $
$\alpha=\frac{5}{3}=1.67$
[Given: The permeability of free space $\mu_0=4 \pi \times 10^{-7} \mathrm{~N} \mathrm{~A}^{-2}$ ]
Explanation:
$ \begin{aligned} &R=0.1 \Omega\\\\ &\varepsilon=\left(\mathrm{B}_1-\mathrm{B}_2\right) b v_{\mathrm{y}}\\\\ &\begin{aligned} & \mathrm{i}=\frac{\varepsilon}{R}=\frac{\mu_0 I}{2 \pi R}\left(\frac{1}{d}-\frac{1}{d+a}\right) b v_y \\\\ & \Rightarrow 10^{-5}=\frac{2 \times 10^{-7} \times 10}{0.1}\left[\frac{1}{4}-\frac{1}{8}\right] \times 2 . v_y \end{aligned} \end{aligned} $
$ \therefore \mathrm{v}_{\mathrm{y}}=2 $
$ \begin{aligned} & \tan \theta=\frac{\mathrm{v}_{\mathrm{y}}}{\mathrm{v}_{\mathrm{x}}}=\frac{1}{\sqrt{3}} \\\\ & \therefore \mathrm{v}_{\mathrm{x}}=2 \sqrt{3} \\\\ & \therefore \mathrm{v}=\sqrt{\mathrm{v}_{\mathrm{x}}^2+\mathrm{v}_{\mathrm{y}}^2}=4 \end{aligned} $
Explanation:
${r_s} = {{\sqrt {2{m_s}{q_s}V} }\over {{q_s}B}}$
${{{r_s}} \over {{r_\alpha }}} = \sqrt {{{{m_s}} \over {{q_s}}}{{{q_\alpha }} \over {{m_\alpha }}}} = \sqrt {\left( {{{32} \over 1}} \right)\left( {{2 \over 4}} \right)} $
${{{r_s}} \over {{r_\alpha }}} = 4$
Explanation:
The magnetic moment of a coil having n turns, area A and current i is given by $\overrightarrow \mu = ni\overrightarrow A $. The torque on this coil when placed in a magnetic field $\overrightarrow B $ is given by $\overrightarrow \tau = \overrightarrow \mu \times \overrightarrow B $. In a moving coil galvanometer, cylindrical magnets ensure that $\overrightarrow A $ and $\overrightarrow B $ are perpendicular to each other. Thus, deflection torque on the coil is $\tau$d = niAB. The restoring torque on a coil (at deflection angle $\theta$) due to the wire of torsional constant k is $\tau$r = k$\theta$. In equilibrium, deflection torque is balanced by the restoring torque i.e., $\tau$d = $\tau$r, which gives
$i = {k \over {nAB}}\theta $.
The maximum deflection current of the galvanometer (ig) occurs at full scale deflection $\theta$max i.e.,
${i_g} = {{k{\theta _{\max }}} \over {nAB}} = {{({{10}^{ - 4}})(0.2)} \over {(50)(2 \times {{10}^{ - 4}})(0.02)}} = 0.01$ A.
A galvanometer of resistance G is converted to an ammeter by connecting a small shunt resistance S in parallel.
Kirchhoff's loop law gives
${i_g}G - (i - {i_g})S = 0$, $ \Rightarrow S = {i_g}G/(i - {i_g})$.
The maximum deflection current of galvanometer sets upper limit on the current measured by this ammeter (imax). Substitute the values to get
$S = {{{i_g}G} \over {{i_{\max }} - {i_g}}} = {{(0.1)(50)} \over {1 - 0.1}} = {5 \over {0.9}} = 5.55\Omega $.
Explanation:
The charged particle enters the magnetic field region at O. Its velocity is perpendicular to the field direction. The magnetic force on the particle, qv0B1, provides centripetal acceleration to move on a circular path from O to P. The radius of the circular path is given by
${r_1} = m{v_0}/(q{B_1})$.
The time taken by the particle to travel from O to P is
${T_1} = \pi {r_1}/{v_0} = \pi m/(q{B_1})$
At P, the particle enters the magnetic field B2 with its velocity perpendicular to the field direction. The magnetic force qv0B2 provides the centripetal acceleration to move in a circular path from P to Q. The radius of the circular path is
${r_2} = m{v_0}/(q{B_2})$.
The time taken by the particle to travel from P to Q is
${T_2} = \pi {r_2}/{v_0} = \pi m/(q{B_2})$.
AT Q, the particle crosses x-axis from below for the first time. Thus, T = T1 + T2. The average speed of the particle along the x-axis in the time interval T is given by
$v = {{\int_0^T {{v_x}dt} } \over T} = {{\int_0^{{T_1}} {{v_{x1}}dt + \int_{{T_1}}^{{T_1} + {T_2}} {{v_{x2}}dt} } } \over {{T_1} + {T_2}}}$
$ = {{2{r_1} + 2{r_2}} \over {{T_1} + {T_2}}} = {{{{2m{v_0}} \over q}\left( {{1 \over {{B_1}}} + {1 \over {{B_2}}}} \right)} \over {{{\pi m} \over q}\left( {{1 \over {{B_1}}} + {1 \over {{B_2}}}} \right)}}$
$ = {{2{v_0}} \over \pi } = {{2(\pi )} \over \pi } = 2$ m/s
Explanation:
When current in wires is in same direction, the magnetic fields due to two wires all in opposite direction.
From $B = {{{\mu _0}} \over {4\pi }}.{{2I} \over r}$, we get
${B_1} = {{{\mu _0}} \over {4\pi }}.2I\left[ {{1 \over {{x_1}}} - {1 \over {({x_0} - {x_1})}}} \right]$
$ = {{{\mu _0}I} \over {2\pi }}\left[ {{{{x_0} - {x_1} - {x_1}} \over {{x_1}({x_0} - {x_1})}}} \right]$
$ = {{{\mu _0}I} \over {2\pi }}\left[ {{{{x_0} - 2{x_1}} \over {{x_1}({x_0} - {x_1})}}} \right]$ ..... (1)
When the direction of current in two wires is opposite, field will be in the same direction.
${B_2} = {{{\mu _0}I} \over {2\pi }}\left[ {{1 \over {{x_1}}} + {1 \over {({x_0} - {x_1})}}} \right]$
${B_2} = {{{\mu _0}I} \over {2\pi }}\left[ {{{{x_0} - {x_1} + {x_1}} \over {{x_1}({x_0} - {x_1})}}} \right]$
${B_2} = {{{\mu _0}I} \over {2\pi }}\left[ {{{{x_0}} \over {{x_1}({x_0} - {x_1})}}} \right]$
From ${{m{v^2}} \over r} = qvB$ or $v = {{qBr} \over m}$ or $r = {{mv} \over {qB}}$
$B \propto {1 \over r}$
Therefore, ${{{R_1}} \over {{R_2}}} = {{{B_2}} \over {{B_1}}} = {{{x_0}} \over {({x_0} - 2{x_1})}}$
${{{R_1}} \over {{R_2}}} = {{{x_0}/{x_1}} \over {({x_0}/{x_1}) - (2{x_1}/{x_1})}} = {3 \over {3 - 2}} = 3$
A cylinder cavity of diameter a exists inside a cylinder of diameter 2a as shown in the figure. Both the cylinder and the cavity are infinitely long. A uniform current density J flows along the length. If the magnitude of the magnetic field at the point P is given by ${N \over {12}}{\mu _0}aJ$, then the value of N is ______________.
Explanation:
BR = BT $-$ BC
R = Remaining portion
T = Total portion and
C = cavity
${B_R} = {{{\mu _0}{I_T}} \over {2a\pi }} - {{{\mu _0}{I_C}} \over {2(3a/2)\pi }}$ ..... (i)
${I_T} = J(\pi {a^2})$
${I_C} = J\left( {{{\pi {a^2}} \over 4}} \right)$
Substituting the values in Eq. (i), we have
${B_R} = {{{\mu _0}} \over {a\pi }}\left[ {{{{I_T}} \over 2} - {{{I_C}} \over 3}} \right]$
$ = {{{\mu _0}} \over {a\pi }}\left[ {{{\pi {a^2}J} \over 2} - {{\pi {a^2}J} \over {12}}} \right]$
$ = {{5{\mu _0}aJ} \over {12}}$
$\therefore$ $N = 5$
A long circular tube of length 10 m and radius 0.3 m carries a current I along its curved surface as shown. A wire-loop of resistance 0.005 $\Omega$ and of radius 0.1 m is placed inside the tube with its axis coinciding with the axis of the tube. The current varies as $I = {I_0}\cos (300t)$, where I0 is constant. If the magnetic moment of the loop is $N{\mu _0}{I_0}\sin (300t)$, then N is ___________.
Explanation:
The flux through the ring is
$\phi = B\pi {r^2}$
Assuming the cylinder as a solenoid with close winding, we have
$B = {{{\mu _0}I} \over L}$
Therefore,
$\phi = \left( {{{{\mu _0}I} \over L}} \right)\pi {r^2}\cos 300t$
The induced emf is
$\varepsilon = {{ - d\phi } \over {dt}} = 300\left( {{{{\mu _0}I} \over L}} \right)\pi {r^2}\sin 300t$
Therefore, the current induced is
$i = {\varepsilon \over R} = \left( {{{\pi {r^2}300} \over {RL}}} \right){\mu _0}{I_0}\sin 300t$
The magnetic moment is
M = Current $\times$ Area of loop
Therefore,
$m = \left( {{{{{(3.14)}^2} \times {{(0.1)}^4} \times 300} \over {0.005 \times 10}}} \right){\mu _0}{I_0}\sin 300t$
$ = 6{\mu _0}{I_0}\sin 300t$
Hence, N = 6.
A steady current I goes through a wire loop PQR having shape of a right angle triangle with PQ = 3x, PR = 4x and QR = 5x. If the magnitude of the magnetic field at P due to this loop is $k\left( {{{{\mu _0}I} \over {48\pi x}}} \right)$, find the value of $k$.
Explanation:
The segments PQ and PR, in the triangle shown here, cannot produce $\overrightarrow B $ at point P since point P lies on them. Here, only QR creates $\overrightarrow B $ at point P.
Applying ${B_P} = {{{\mu _0}i} \over {4\pi R}}(\cos {\phi _1} + \cos {\phi _2})$, where ${\phi _1} = 53^\circ $ and ${\phi _2} = 37^\circ $, we get the magnitude of the magnetic field as follows:
${B_P} = {{{\mu _0}I} \over {4\pi (4\pi \sin 37^\circ )}}(\cos 53^\circ + \cos 37^\circ )$
$ = {{{\mu _0}I} \over {16\pi x(3/5)}}\left( {{{3x} \over {5x}} + {{4x} \over {5x}}} \right)$
$ = {{5{\mu _0}I} \over {48\pi x}}\left( {{7 \over 5}} \right) = {{7{\mu _0}I} \over {48\pi x}} = k\left( {{{{\mu _0}I} \over {48\pi x}}} \right)$
where $k = 7$.
A thin stiff insulated metal wire is bent into a circular loop with its two ends extending tangentially from the same point of the loop. The wire loop has mass $m$ and radius $r$ and it is in a uniform vertical magnetic field $B_0$, as shown in the figure. Initially, it hangs vertically downwards, because of acceleration due to gravity $g$, on two conducting supports at $\mathrm{P}$ and $\mathrm{Q}$. When a current $I$ is passed through the loop, the loop turns about the line $\mathrm{PQ}$ by an angle $\theta$ given by
An infinitely long wire, located on the $z$-axis, carries a current $I$ along the $+z$-direction and produces the magnetic field $\vec{B}$. The magnitude of the line integral $\int \vec{B} \cdot \overrightarrow{d l}$ along a straight line from the point $(-\sqrt{3} a, a, 0)$ to $(a, a, 0)$ is given by
[ $\mu_0$ is the magnetic permeability of free space.]
Which one of the following options represents the magnetic field $\vec{B}$ at $\mathrm{O}$ due to the current flowing in the given wire segments lying on the $x y$ plane?
A small circular loop of area $A$ and resistance $R$ is fixed on a horizontal $x y$-plane with the center of the loop always on the axis $\hat{n}$ of a long solenoid. The solenoid has $m$ turns per unit length and carries current $I$ counterclockwise as shown in the figure. The magnetic field due to the solenoid is in $\hat{n}$ direction. List-I gives time dependences of $\hat{n}$ in terms of a constant angular frequency $\omega$. List-II gives the torques experienced by the circular loop at time $t=\frac{\pi}{6 \omega}$. Let $\alpha=\frac{A^{2} \mu_{0}^{2} m^{2} I^{2} \omega}{2 R}$.
| List-I | List-II |
|---|---|
| (I) $\frac{1}{\sqrt{2}}(\sin \omega t \hat{\jmath}+\cos \omega t \hat{k})$ | (P) 0 |
| (II) $\frac{1}{\sqrt{2}}(\sin \omega t \hat{\imath}+\cos \omega t \hat{\jmath})$ | (Q) $-\frac{\alpha}{4} \hat{\imath}$ |
| (III) $\frac{1}{\sqrt{2}}(\sin \omega t \hat{\imath}+\cos \omega t \hat{k})$ | (R) $\frac{3 \alpha}{4} \hat{\imath}$ |
| (IV) $\frac{1}{\sqrt{2}}(\cos \omega t \hat{\jmath}+\sin \omega t \hat{k})$ | (S) $\frac{\alpha}{4} \hat{\jmath}$ |
| (T) $-\frac{3 \alpha}{4} \hat{\imath}$ |
Which one of the following options is correct?
When d $\approx$ a but wires are not touching the loop, it is found that the net magnetic field on the axis of the loop is zero at a height h above the loop. In that case
Consider d >> a, and the loop is rotated about its diameter parallel to the wires by 30$^\circ$ from the position shown in the below figure. If the currents in the wires are in the opposite directions, the torque on the loop at its new position will be (assume that the net field due to the wires is constant over the loop)
The change in the magnetic dipole moment associated with the orbit, at the end of the time interval of the magnetic field change, is
A loop carrying current $l$ lies in the xy-plane as shown in the figure. The unit vector $\widehat k$ is coming out of the plane of the paper. The magnetic moment of the current loop is
An infinite long hollow conducting cylinder with inner radius R/2 and outer radius R carries a uniform current density along its length. The magnitude of the magnetic field, $\left| {\overrightarrow B } \right|$ as a function of the radial distance r from the axis is best represented by
Estimate the wavelength at which plasma reflection will occur for a metal having the density of electrons N $ \approx $ 4 $ \times $ 1027 m-3. Taking ${{\varepsilon _0}}$ = 10- 11 and m $ \approx $ 10- 30, where these quantities are in proper SI units.
A long insulated copper wire is closely wound as a spiral of N turns. The spiral has inner radius a and outer radius b. The spiral lies in the xy-plane and a steady current I flows through the wire. The z-component of the magnetic field at the centre of the spiral is
A thin flexible wire of length L is connected to two adjacent fixed points and carries a current I in the clockwise direction, as shown in the figure. When the system is put in a uniform magnetic field of strength B going into the plane of the paper, the wire takes the shape of a circle. The tension in the wire is
In the graph below, the resistance R of a superconductor is shown as a friction of its temperature T for two different magnetic fields B1 (solid line) and B2 (dashed line). If B2 is larger than B1 which of the following graphs shows the correct variation of R with T in these fields?
A superconductor has Tc(0) = 100 K. When a magnetic field of 7.5 T is applied, its Tc decreases to 75 K. For this material, one can definitely say that when
STATEMENT 1 : The sensitivity of a moving coil galvanometer is increased by placing a suitable magnetic material as a core inside the coil.
and
STATEMENT 2 : Soft iron has a high magnetic permeability and cannot be easily magnetized or demagnetized.
A magnetic field $\overrightarrow{\mathrm{B}}=\mathrm{B}_{0} \hat{j}$ exists in the region $a < x < 2 a$ and $\overrightarrow{\mathrm{B}}=-\mathrm{B}_{0} \hat{j}$, in the region $2 a < x < 3 a$, where $\mathrm{B}_{0}$ is a positive constant. A positive point charge moving with a velocity $\vec{v}=v_{0} \hat{i}$, where $v_{0}$ is a positive constant, enters the magnetic field at $x=a$. The trajectory of the charge in this region can be like,
Two wires each carrying a steady current I are shown in four configurations in Column I. Some of the resulting effects are described in Column II. Match the statements in Column I with the statements in Column II and indicate your answer by darkening appropriate bubbles in the $4 \times 4$ matrix given in the ORS.
| Column I | Column II | ||
|---|---|---|---|
| (A) | Point P is situated midway between the wires. |
(P) | The magnetic fields (B) at P due to the currents in the wire are in same direction. |
| (B) | Point P is situated at the mid-point of the line joining the centers of the circular wires, which have same radii. |
(Q) | The magnetic fields (B) at P due to the currents in the wires are in opposite directions. |
| (C) | Point P is situated at the mid-point of the line joining the centers of the circular wires, which have same radii. |
(R) | There is no magnetic field at P. |
| (D) | Point P is situated at the common center of the wires. |
(S) | The wires repel each other. |
In a moving coil galvanometer, torque on the coil can be expressed as $\tau=k i$, where $i$ is current through the wire and $k$ is constant. The rectangular coil of the galvanometer having numbers of turns $\mathrm{N}$, area $\mathrm{A}$ and moment of inertia I is placed in magnetic field B. Find
(A) $k$ in terms of given parameters $\mathrm{N}, \mathrm{I}, \mathrm{A}$ and B;
(B) The torsional constant of the spring, if a current $i_{0}$ produces a deflection of $\frac{\pi}{2}$ in the coil;
(C) The maximum angle through which coil is deflected, if charge $\mathrm{Q}$ is passed through the coil almost instantaneously. (Ignore the damping in mechanical oscillations.)
A positive, singly ionized atom of mass number $A_{\mathrm{M}}$ is accelerated from rest by the voltage $192 \mathrm{~V}$. Thereafter, it enters a rectangular region of width $w$ with magnetic field $\vec{B}_0=0.1 \hat{k}$ Tesla, as shown in the figure. The ion finally hits a detector at the distance $x$ below its starting trajectory.
[Given: Mass of neutron/proton $=(5 / 3) \times 10^{-27} \mathrm{~kg}$, charge of the electron $=1.6 \times 10^{-19} \mathrm{C}$.]
Which of the following option(s) is(are) correct?
A resistor R = 1.4 $\Omega$ and a capacitor C0 = 5.0$\mu$F are connected in series between the rails. At time t = 0, C0 is uncharged. Which of the following statement(s) is(are) correct? [$\mu$0 = 4$\pi$ $\times$ 10$-$7 SI units. Take ln 2 = 0.7]
A steady current I flows along an infinitely long hollow cylindrical conductor of radius R. This cylinder is placed coaxially inside an infinite solenoid of radius 2R. The solenoid has n turns per unit length and carries a steady current I. Consider a point P at a distance r from the common axis. The correct statement(s) is(are)
A particle of mass M and positive charge Q, moving with a constant velocity ${\overrightarrow u _1} = 4\widehat i$ ms$-$1 enters a region of uniform static magnetic field, normal to the xy plane. The region of the magnetic field extends from x = 0 to x = L for all values of y. After passing through this region, the particle emerges on the other side after 10 ms with a velocity ${\overrightarrow u _2} = 2\left( {\sqrt 3 \widehat i + \widehat j} \right)$ ms$-$1. The correct statement(s) is(are)
Consider the motion of a positive point charge in a region, there are simultaneous uniform electric and magnetic fields $\overrightarrow E = {E_0}\widehat j$ and $\overrightarrow B = {B_0}\widehat j$. At time t = 0, this charge has velocity $\overrightarrow v $ in the xy-plane, making an angle $\theta$ with the x-axis. Which of the following option(s) is(are) correct for time t > 0 ?
An electron and a proton are moving on straight parallel paths with same velocity. They enter a semi-infinite region of uniform magnetic field perpendicular to the velocity. Which of the following statement(s) is/are true?
A particle of mass m and charge q, moving with velocity v enters Region II normal to the boundary as shown in the figure. Region II has a uniform magnetic field B perpendicular to the plane of the paper. The length of the Region II is $l$. Choose the correct choice (s).
An infinite current-carrying wire passes through point O and in perpendicular to the plane containing a current-carrying loop ABCD as shown in the figure. Choose the correct option(s):
Net force on the loop is zero.
Net torque on the loop is zero.
As seen from O , the loop rotates clockwise.
As seen from O , the loop rotates anticlockwise.