Magnetism

Let loop makes angle $\theta$ with vertical.

in equilibrium $\tau_{\text {net }}=0$

$\tau_0=\mathrm{MB} \sin (90-\theta)-\mathrm{mg} . \mathrm{r} \sin \theta=0$

$\mathrm{I}$. $\pi \mathrm{r}^2 \cdot \mathrm{B}_0 \cos \theta=\mathrm{mg} \mathrm{r} \cdot \sin \theta$

$\tan \theta=\frac{\pi \mathrm{rIB}_0}{\mathrm{mg}}$

$\begin{aligned} & \Rightarrow|\overrightarrow{\mathrm{d} \ell}|=\mathrm{rd} \theta \\ & \Rightarrow|\overrightarrow{\mathrm{B}}|=\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{r}} \\ & \Rightarrow \int \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d} \ell}=\int|\overline{\mathrm{B}}||\overrightarrow{\mathrm{d} \ell}| \cos 0^{\circ} \\ & =\int\left(\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{r}}\right) \times(\mathrm{rd} \theta) \\ & =\int_\limits{\theta_1}^{\theta_2} \frac{\mu_0 \mathrm{I}}{2 \pi} \mathrm{d} \theta=\frac{\mu_0 \mathrm{I}}{2 \pi}\left[\theta_2-\left(-\theta_1\right)\right] \end{aligned}$

$\text { [ } \theta_1 \text { is anticlockwise hence taken negative] }$

$\begin{aligned} & \Rightarrow \tan \theta_1=\frac{a \sqrt{3}}{a}=\sqrt{3} \\ & \theta_1=\frac{\pi}{3} \\ & \Rightarrow \tan \theta_2=\frac{a}{a}=1 \\ & \theta_2=\frac{\pi}{4} \\ & \Rightarrow \int B \cdot d \ell=\frac{\mu_0 I}{2 \pi}\left[\frac{\pi}{3}+\frac{\pi}{4}\right] \\ & =\frac{7 \mu_0 I}{24} \\ & \Rightarrow \text { Ans. Option (A) } \end{aligned}$

$B = {{{\mu _0}i} \over {4\pi r}}\left( {\sin \alpha + \sin \beta } \right)$

Direction of B :

Note :-

On the axis of the wire magnetic field = 0.

Case 1 :

Infinite long wire (Very Long wire) :

$\alpha ,\beta \to 90^\circ $

$\therefore$ $B = {{{\mu _0}i} \over {4\pi r}}(\sin 90^\circ + \sin 90^\circ )$

$ = {{{\mu _0}i} \over {4\pi r}}(1 + 1)$

$ = {{{\mu _0}i} \over {2\pi r}}$

Graph :

Case 2 :

One end is at infinite and other end is in front.

Here $\alpha=90^\circ$ and $\beta=0^\circ$

$B = {{{\mu _0}i} \over {4\pi r}}(\sin 90^\circ + \sin 0^\circ )$

$ = {{{\mu _0}i} \over {4\pi r}}$

Case 3 :

When drawn perpendicular from P don't lie on the wire

Here, magnetic field at point P is,

$B = {{{\mu _0}i} \over {4\pi r}}(\sin\alpha-\sin\beta)$

(2) Magnetic Field Due to Half Loop or Semicircle :

Half loop means ${1 \over 2}$ turn. So, N = ${1 \over 2}$.

Magnetic field at centre O,

$\therefore$ B at centre $ = {{{\mu _0} \times {1 \over 2} \times i} \over {2R}} = {{{\mu _0}i} \over {4R}}$

Here,

Here, $N=\frac{1}{2}$

i = Amount of current flowing through loop

R = Radius of half loop

Direction of B found by right hand thumb rule. Curl your right hand's finger in the direction of current then thumb gives direction of magnetic field on axis as well as entire plane.

(3) Magnetic Field Due to Quarter Loop :

Quarter loop means ${1 \over 4}$ turn. So, N = ${1 \over 4 }$.

Magnetic field at O,

$\therefore$ B at centre $ = {{{\mu _0} \times {1 \over 4} \times i} \over {2R}} = {{{\mu _0}i} \over {8R}}$

Here,

Here, $N=\frac{1}{4}$

i = Amount of current flowing through loop

R = Radius of quarter loop

Direction of B found by right hand thumb rule. Curl your right hand's finger in the direction of current then thumb gives direction of magnetic field on axis as well as entire plane.

I. $\phi = B\,A\,\widehat k\,.\,\widehat n$

$\phi = {{BA} \over {\sqrt 2 }}\cos (\omega t)$

$\varepsilon = {{BA\omega } \over {\sqrt 2 }}\sin (\omega t)$

$i = {{BA\omega } \over {\sqrt 2 R}}\sin (\omega t)$

$\overrightarrow m = iA\,\widehat k = {{B{A^2}\omega } \over {\sqrt 2 R}}\sin (\omega t)\widehat k$

$\overrightarrow \tau = \overrightarrow m \times \overrightarrow B = {{{B^2}{A^2}\omega } \over {\sqrt 2 R}}\sin (\omega t)(\widehat k \times \widehat n)$

$ = - {{{B^2}{A^2}\omega } \over {2R}}\left[ {\widehat i} \right]{\sin ^2}(\omega t)$

$\tau = - {{{B^2}{A^2}\omega } \over {2R}}\left[ {{{\sin }^2}\left( {{\pi \over 6}} \right)} \right] = {{ - \alpha } \over 4}\widehat i$

(I) $\to$ Q.

II. $\phi = O$

(II) $\to$ P.

III. $\phi = {{BA\omega } \over {\sqrt 2 }}\cos (\omega t)$

$i = {{BA\omega } \over {\sqrt 2 R}}\sin (\omega t)$

$\overrightarrow m = {{B{A^2}\omega } \over {\sqrt 2 R}}\sin (\omega t)\widehat k$

$\overrightarrow \tau = \overrightarrow m \times \overrightarrow B = {{{B^2}{A^2}\omega } \over {\sqrt 2 \times \sqrt 2 R}}\sin \omega t(\widehat k \times (\sin \omega t\widehat i + \cos \omega t\widehat k))$

$\tau = {{{B^2}{A^2}\omega \sin (\omega t)} \over {2R}}\sin (\omega t)\widehat j$

$ = {{{B^2}{A^2}\omega } \over {2R}}{\sin ^2}(\omega t)\widehat j$

$ = {\alpha \over 4}\widehat j$

(III) $\to$ S.

IV. $\phi = {{BA} \over {\sqrt 2 }}\sin (\omega t)$

$i = - {{BA\omega } \over {\sqrt 2 R}}\cos (\omega t)$

$\overrightarrow m = - {{B{A^2}\omega } \over {\sqrt 2 R}}\cos \omega t(\widehat k)$

$\overrightarrow \tau = \overrightarrow m \times \overrightarrow B = - {{{B^2}{A^2}\omega } \over {2R}}(\widehat k \times \widehat j){\cos ^2}(\omega t)$

$\tau = + {{{B^2}{A^2}\omega } \over {2R}}(\widehat i)\,.\,{\cos ^2}\left( {{\pi \over 6}} \right)$

$ = + {3 \over 4}\alpha \widehat i$

(IV) $\to$ R.

The star shape is composed of 12 wires. Thus, the total magnetic field at centre is 12 times of magnetic field due to one wire.

Let us first calculate the magnetic field due to element AB. Perpendicular distance of centre O from element AB, OP = a

Angle subtended by centre at element are $\theta$₁ = 30$^\circ$, $\theta$₂ = 60$^\circ$ as shown in figure.

$\therefore$ $\overrightarrow B = {{{\mu _0}I} \over {4\pi a}}(\cos 30^\circ - \cos 60^\circ ) \odot $

$ = {{{\mu _0}I} \over {4\pi a}}\left( {{{\sqrt 3 } \over 2} - {1 \over 2}} \right) \odot = {{{\mu _0}I} \over {8\pi a}}(\sqrt 3 - 1) \odot $

Since the direction of magnetic field due to each element side is out of the paper

$\therefore$ Net magnetic field at O $ = 12 \times {{{\mu _0}I} \over {8\pi a}}(\sqrt 3 - 1)$

$ = {{{\mu _0}I} \over {4\pi a}}6(\sqrt 3 - 1)$

The particle will describe a helical path with axis along +z direction if (i) component of net force in x-y plane provides the centripetal acceleration and (ii) the particle has a non-zero velocity and/or non-zero force along +z direction. A proton with initial velocity $\overrightarrow v = 2{{{E_0}} \over {{B_0}}}\widehat x$ will move in a helical path with axis along +z direction when placed in an electric field $\overrightarrow E = {E_0}\widehat z$ and magnetic field $\overrightarrow B = {B_0}\widehat z$.

Lorentz force on a charged particle having a charge q, moving with a velocity $\overrightarrow v $ in a region of uniform electric field $\overrightarrow E $ and uniform magnetic field $\overrightarrow B $, is given by

${\overrightarrow F _{net}} = {\overrightarrow F _E} + {\overrightarrow F _B} = q\overrightarrow E + q\overrightarrow v \times \overrightarrow B $. ...... (1)

The particle will move along $ - \widehat y$ axis if (i) initial velocity $\overrightarrow v $ is along $-\widehat y$ and net force is either zero or directed along $-\widehat y$ (ii) initial velocity $\overrightarrow v$ is zero and net force is directed along $-\widehat y$. A proton with initial velocity $\overrightarrow v = \overrightarrow 0 $ placed in an electric field $\overrightarrow E = - {E_0}\widehat y$ and magnetic field $\overrightarrow B = - {B_0}\widehat y$ will experience a net force ${\overrightarrow F _{net}} = - q{E_0}\widehat y$. It moves in a straight line along $ - \widehat y$.

$\overrightarrow F $ = q$\overrightarrow E $ + q($\overrightarrow v $ $\times$ $\overrightarrow B $) ....... (1)

For a particle to move in a straight line with constant velocity, $\overrightarrow F $ = 0. Therefore, from Eq. (1), we get

$q\overrightarrow E + q(\overrightarrow V \times \overrightarrow B ) = 0$

$ \Rightarrow q\overrightarrow E - q(\overrightarrow v \times \overrightarrow B )$

$ \Rightarrow \overrightarrow E = - (\overrightarrow v \times \overrightarrow B )$

For $\overrightarrow v = {{{E_0}} \over {{B_0}}}\widehat y$ and $\overrightarrow B = {B_0}\widehat z$, we get

$\overrightarrow v \times \overrightarrow B = \left( {{{{E_0}} \over B}\widehat y} \right) \times ({B_0}\widehat z) = {{{E_0}} \over {{B_0}}} \times {B_0}(\widehat y \times z)$

Therefore,

$\overrightarrow v \times \overrightarrow B = {E_0}\widehat x$ (as y $\times$ z = x)

and

$\overrightarrow E = - {E_0}\widehat x$

Therefore,

$v = {{{E_0}} \over {{B_0}}}\widehat y$; $\overrightarrow E = - {E_0}\widehat x$; $\overrightarrow B = {B_0}\widehat z$

Hence, option (D) is correct.

B_R = B due to ring

B₁ = B due to wire-1

B₂ = B due to wire-2

In magnitudes

${B_1} = {B_2} = {{{\mu _0}I} \over {2\pi r}}$

Resultant of B₁ and B₂

$ = 2{B_1}\cos \theta = 2\left( {{{{\mu _0}I} \over {2\pi r}}} \right)\left( {{h \over r}} \right)$

$ = {{{\mu _0}Ih} \over {\pi {r^2}}}$

${B_R} = {{{\mu _0}I{R^2}} \over {2{{({R^2} + {x^2})}^{3/2}}}}$

$ = {{2{\mu _0}I\pi {a^2}} \over {4\pi {r^3}}}$

As, R = a, x = h and a² + h² = r²

For zero magnetic field at P,

${{{\mu _0}Ih} \over {\pi {r^2}}} = {{2{\mu _0}I\pi {a^2}} \over {4\pi {r^3}}}$

$ \Rightarrow \pi {a^2} = 2rh \Rightarrow h \approx 1.2a$

Magnetic field at mid-point of two wires

= 2 (magnetic field due to one wire)

$ = 2\left[ {{{{\mu _0}} \over {2\pi }}{1 \over d}} \right] = {{{\mu _0}I} \over {\pi d}} \otimes $

Magnetic moment of loop

M = IA = I $\pi$a²

Torque on loop = MB sin 150$^\circ$ = ${{{\mu _0}{I^2}{a^2}} \over {2d}}$

Let the point charge Q is moving in a circle of constant radius R in anticlockwise direction as shown in the figure.

The magnetic flux through the circular loop at time t is given by $\phi = B(\overrightarrow t )\,.\,(\pi {R^2}\widehat z)\, = \pi B(t){R^2}$. Faraday's law gives the induced emf e as

$e = - {{d\phi } \over {dt}} = - \pi {R^2}{{dB(t)} \over {dt}} = - \pi {R^2}B$ ..... (1)

Lenz's law gives the direction of induced current and hence electric field $(\overrightarrow E )$ as clockwise. The induced emf is related to $\overrightarrow E $ by

$e = \oint {\overrightarrow E .\,d\overrightarrow l = - E(2\pi R)} $ ..... (2)

Eliminate e from equations (1) and (2) to get E = BR/2.

${M \over L} = {Q \over {2m}}$

$\therefore$ $M = \left( {{Q \over {2m}}} \right)L$

$ \Rightarrow M \propto L$, where $\gamma = {Q \over {2m}}$

$\left( {{Q \over {2m}}} \right)(I\omega )$

$ = \left( {{Q \over {2m}}} \right)(m{R^2}\omega ) = {{Q\omega {R^2}} \over 2}$

Induced electric field is opposite. Therefore,

$\omega ' = \omega - \alpha t$

$\alpha = {\tau \over I} = {{(QE)R} \over {m{R^2}}} = {{(Q)\left( {{{BR} \over 2}} \right)R} \over {m{R^2}}} = {{QB} \over {2m}}$

$\therefore$ $\omega ' = \omega - {{QB} \over {2m}}.1 = \omega - {{QB} \over {2m}}$

${M_f} = {{Q\omega '{R^2}} \over 2} = Q\left( {\omega - {{QB} \over {2m}}} \right){{{R^2}} \over 2}$

$\therefore$ $\Delta M = {M_f} - {M_i} = - {{{Q^2}B{R^2}} \over {4m}}$

$M = - \gamma {{QB{R^2}} \over 2}$ (as $\gamma = {Q \over {2m}}$)

Area of the loop

$A = \left[ {{a^2} + 4 \times {{\pi {{\left( {{a \over 2}} \right)}^2}} \over 2}} \right]\widehat k = \left[ {{a^2} + {{\pi {a^2}} \over 2}} \right]\widehat k$

Therefore, the magnetic moment of the current loop is

$M = I \times A = I\left[ {{a^2} + {{\pi {a^2}} \over 2}} \right]\widehat k = \left[ {1 + {\pi \over 2}} \right]I{a^2}\widehat k$

r = distance of a point from centre.

For r $\le$ R/2 Using Ampere's circuital law,

$\oint {B.\,dl} $

or, $Bl = {\mu _0}({I_{in}})$

$B(2\pi r) = {\mu _0}({I_{in}})$

$B = {{{\mu _0}} \over {2\pi }}{{{I_{in}}} \over r}$ ...... (i)

Since, ${I_{in}} = 0$ $\therefore$ $B = 0$

For ${R \over 2} \le r \le R$

${I_{in}} = \left[ {\pi {r^2} - \pi {{\left( {{R \over 2}} \right)}^2}} \right]\sigma $

Here $\sigma$ = current per unit area.

Substituting in Eq. (i), we have

$B = {{{\mu _0}} \over {2\pi }}{{\left[ {\pi {r^2} - \pi {{{R^2}} \over 2}} \right]\sigma } \over r}$

$ = {{{\mu _0}\sigma } \over {2r}}\left( {{r^2} - {{{R^2}} \over 4}} \right)$

At $r = {R \over 2},\,B = 0$

At $r = R,\,B = {{3{\mu _0}\sigma R} \over 8}$

For r $\ge$ R

${I_{in}} = {I_{Total}} = I$ (say)

Therefore, substituting in Eq. (i), we have

$B = {{{\mu _0}} \over {2\pi }}.\,{I \over r}$

or, $B \propto {1 \over r}$

For resonance

$\omega = {\omega _p} = \sqrt {{{N{e^2}} \over {m{\varepsilon _0}}}} $ [From previous question]

$\upsilon = {\omega \over {2\pi }} = {1 \over {2\pi }}\sqrt {{{N{e^2}} \over {m{\varepsilon _0}}}} $

$\lambda = {c \over \upsilon } = 2\pi c\sqrt {{{m{\varepsilon _0}} \over {N{e^2}}}} $

Substituting the given values, we get

$\lambda = 2 \times 3.14 \times 3 \times {10^8} \times \sqrt {{{{{10}^{ - 30}} \times {{10}^{ - 11}}} \over {4 \times {{10}^{27}} \times {{(1.6 \times {{10}^{ - 19}})}^2}}}} $

$ = {{2 \times 3.14 \times 3 \times {{10}^8}} \over {1.6 \times {{10}^{ - 19}}}}\sqrt {{{{{10}^{ - 41}}} \over {4 \times {{10}^{27}}}}} = {{9.42} \over {1.6}} \times {10^{27}} \times {10^{ - 34}}$ m

$ \approx 6.00 \times {10^{ - 7}}$ m = 600 nm

Magnetic field at the centre of a circular loop of radius r and carrying a current $I = {{{\mu _0}I} \over {2r}}$. The direction of the field is along z-direction if the current is anticlockwise.

Consider a small element of width dr. The current through the element is

$dI = {{total\,current\,in\,spiral} \over {total\,width\,of\,spiral}} \times width\,of\,element$

$ = {{Idr} \over {(b - a)}}$

$\therefore$ $B = \int\limits_a^b {{{{\mu _0}NdI} \over {2r}} = \int\limits_a^b {{{{\mu _0}NI} \over {2(b - a)}}{{dr} \over r}} } $

$ = {{{\mu _0}NI} \over {2(b - a)}}\int\limits_a^b {{{dr} \over r} = {{{\mu _0}NI} \over {2(b - a)}}\ln \left( {{b \over a}} \right)} $

Consider an small element AB of length dl of the circle of radius R subtending an angle $\theta$ at the centre O.

If T is the tension in the wire, then force towards the centre will be equal to $2T\sin \left( {{\theta \over 2}} \right)$ which is balanced by outward magnetic force on the current carrying element $( = IdlB)$

$2T\sin \left( {{\theta \over 2}} \right) = IdlB$

For small angle $\theta$, $\sin {\theta \over 2} \approx {\theta \over 2}$

or, $T = {{IBdl} \over \theta } = IBR$ ($\because$ $\theta = {{dl} \over R}$)

$ = {{IBL} \over {2\pi }}$ ($\because$ $R = {L \over {2\pi }}$)

From the given figure, T_c(B) is a monotonically decreasing function of B. Thus, B₂ > B₁ implies T_c(B₂) < T_c(B₁). Hence resistance with B₂ will become zero at lower temperature in comparison to B₁.

It is given that T_c(0) = 100 K and T_c(7.5) = 75 K. Since T_c(B) is a monotonically decreasing function of B, T_c(5) < T_c(0) and T_c(5) > T_c(7.5). Thus, 75 K < T_c(5) < 100 K.

Statement 1 is true.

We know that,

Sensitivity $ = {\theta \over I} = {{NBA} \over C}$

If we place soft iron in magnetic field it get magnetise so magnetic field increases. Hence, sensitivity of a moving coil will also increase.

Therefore, $\overrightarrow B $ the ${\theta \over I}$ also increases.

When we place any suitable magnetic material inside the core, the magnetic field will increase.

Statement 2 is wrong because soft iron can be easily magnetized and de magnetized.

For $a < x < 2 a$

$\overrightarrow{\mathrm{B}}=\mathrm{B}_{0} \hat{j} $

The initial velocity is $\vec{v}=v \cdot \hat{j}$

From the above diagram it is clear that force on the particle is toward $+z$-axis (by applying Fleming's left hand rule) at $x=a$, which shifts the particle as shown in the x-z plane

For $2 a < x < 3 a$

$\overrightarrow{\mathrm{B}}=-\mathrm{B}_{0} \hat{j}$

The direction of velocity is shown at $x=2 a$.

Again by using Fleming's left hand rule, we get direction of force.

We can find the direction of magnetic field at $P$ due to a wire by using thumb rule, and the direction of force on a wire due to other is calculated by the direction of cross product of direction of current and direction of magnetic field, $\vec{f}=q(\vec{l} \times \vec{B})$

(A) From thumb rule, the magnetic field at point $\mathrm{P}$ due to both wires is in opposite direction. Since $\mathrm{P}$ is at midpoint, net magnetic field at $P$ is zero. The direction of force on one wire is towards another (from above equation), hence they attract each other. Thus, option $(\mathrm{Q})$ are $(\mathrm{R})$ are correct.

(B) Direction of magnetic field at $\mathrm{P}$ is same due to both the loop. Hence option $(\mathrm{P})$ is correct. If both the loop is very close they might act as single (like solenoid), so we cannot predict loop will attract nor repel.

(C) Similar to case $A$, net $B$ at point $\mathrm{P}$ is zero. And direction of force is towards another wire which means attraction. Hence, option (Q) and (R) are correct.

(D) Magnetic field at the center of current carrying loop is given by $\mathrm{B}=\frac{\propto_{0} i}{2 R}$

Current in the two loop is in opposite direction so magnetic field is in opposite direction due to both the loops. Radius of two loops is different, so it will have non zero field at point $P$.

(A) The question is based on the theory of a moving coil galvanometer.

Torque on Coil

$\begin{aligned} \tau & =\mathrm{MB} \sin \theta \\ \tau & =(\mathrm{N} i \mathrm{~A}) \mathrm{B} \sin 90^{\circ} \\ \tau & =\mathrm{N} i \mathrm{AB} \\ k i & =\mathrm{NiAB} \\ k & =\mathrm{NAB} \end{aligned}$

(B) The torsional constant of the spring is defined as the torque per unit twist. Thus $\tau= \mathrm{C} \theta$.

But

$iNAB = C\theta $

Given, $i = {i_0}$ and $\theta = \pi /2$

${i_0}NAB = C{\pi \over 2}$

Hence, the torsional constant of the spring

$C = {{2N{i_0}AB} \over \pi }$

(C) Angular impulse $=\int \tau d t=\int \mathrm{NiAB} d t$

$=\mathrm{NAB} \int i d t=\mathrm{NABQ}$

But, I $\omega=\mathrm{NABQ}$ (or) $\omega=\frac{\text { NABQ }}{\mathrm{I}}$

$\therefore$ Rotational kinetic energy

$\begin{aligned} & =\frac{1}{2} \mathrm{I} \omega^{2}=\frac{1}{2} \mathrm{I}\left(\frac{\mathrm{NABQ}}{\mathrm{I}}\right)^{2} \\ & =\frac{1}{2} \times \frac{\mathrm{N}^{2} A^{2} \mathrm{~B}^{2} Q^{2}}{\mathrm{I}} \end{aligned}$

$\therefore$ Potential energy $=\frac{1}{2} \mathrm{CQ}_{\max }^{2}$

On equating both Kinetic and Potential energy

$\begin{aligned} \frac{1}{2} \mathrm{Q}_{\max }^{2} \mathrm{C} & =\frac{1}{2} \times \frac{\mathrm{N}^{2} \mathrm{~A}^{2} \mathrm{~B}^{2} \mathrm{Q}^{2}}{\mathrm{I}} \\ \mathrm{Q}_{\max }^{2} & =\frac{\mathrm{N}^{2} \mathrm{~A}^{2} \mathrm{~B}^{2} a^{2}}{\mathrm{I}} \times \frac{1}{\mathrm{C}} \\ & =\frac{\mathrm{N}^{2} \mathrm{~A}^{2} \mathrm{~B}^{2} \mathrm{Q}^{2}}{\mathrm{I}} \times \frac{\pi}{2 i_{0} \mathrm{NAB}} \\ & =\frac{\mathrm{Q}^{2} \mathrm{NAB} \pi}{2 \mathrm{Ii}_{0}} \\ \mathrm{Q}_{\max } & =\sqrt{\frac{\mathrm{Q}^{2} \mathrm{NAB} \pi}{2 \mathrm{Ii}_{0}}}=\mathrm{Q} \sqrt{\frac{\mathrm{NAB} \pi}{2 \mathrm{Ii}_{0}}} \end{aligned}$