Magnetism

$\begin{aligned} & x=2 R \\ & \Rightarrow x=2 \frac{P}{q B} \Rightarrow x=\frac{2 \sqrt{2 m q V}}{q B} \Rightarrow x=\frac{2}{B} \sqrt{\frac{2 m V}{q}} \end{aligned}$

Option A

For $\mathrm{H}^{+} \rightarrow \mathrm{m}=\frac{5}{3} \times 10^{-27} \mathrm{~kg}$

$\therefore x=\frac{2}{0.1} \sqrt{\frac{2 \times \frac{5}{3} \times 10^{-27} \times 192}{1.6 \times 10^{-19}}}=4 \mathrm{~cm}$

Option B

For $\mathrm{A}_{\mathrm{m}}=144$

$x=\frac{2}{0.1} \sqrt{\frac{2 \times 144 \times \frac{5}{3} \times 10^{-27} \times 192}{1.6 \times 10^{-19}}}=48 \mathrm{~cm}$

Option C

for $\mathrm{A}_{\mathrm{m}}=1$

$\mathrm{x}=4 \mathrm{~cm}$ & for $\mathrm{A}_{\mathrm{m}}=196$

$\mathrm{x}=56 \mathrm{~cm}$.

so $\mathrm{x}_0=4 \mathrm{~cm}$ & $\mathrm{x}_1=56 \mathrm{~cm}$

$\therefore \mathrm{x}_1-\mathrm{x}_0=52 \mathrm{~cm}$.

Option D

Minimum width $=\mathrm{R}$

for $\mathrm{A}_M=196$

$\mathrm{R}=\frac{\mathrm{P}}{\mathrm{qB}}=\frac{\sqrt{2 \mathrm{mqV}}}{\mathrm{qB}}$

$\mathrm{R}=\frac{1}{\mathrm{~B}} \sqrt{\frac{2 \mathrm{mV}}{\mathrm{q}}}$

$\mathrm{w}_{\min }=\mathrm{R}=\frac{1}{0.1} \sqrt{\frac{2 \times 196 \times \frac{5}{3} \times 10^{-27} \times 192}{1.6 \times 10^{-19}}}=28 \mathrm{~cm}$

(a) Magnetic field at the plane of the ring is perpendicular to the plane. However, they bend as they move forward.


(b) By symmetry, we can say that B will be same at all the points having the same radial distance. So, B (x, y) will depend on the radial distance $r = \sqrt {{x^2} + {y^2}} $.


(c) ${({B_{net}})_{centre}} = {{{\mu _0}{I_2}} \over {2(2R)}} - {{{\mu _0}{I_1}} \over {2R}}$

$ = {{{\mu _0}} \over {4R}}({I_2} - 2{I_1})$, since I₂ > 2I₁, so B_net at the centre will be non-zero in $ \otimes $ direction. But at some other point, B_net may be zero.

From the graph, it is clear that B_net = 0 for r$\in$(0, R).

So, option (c) is incorrect.


(d) For the graph, it is clear that B = $-$ ve

In, $ \otimes $ direction for r $\in$ (R to 2R), so option (d) is also incorrect.

Let a small element of length $d x$ be chosen at a distance $x$ from the wire. Induced emf de across it is given by

$ \begin{aligned} d \mathrm{E} & =\mathrm{B} v(d x) \\\\ & =\frac{\mu_0 \mathrm{I}}{2 \pi x} v(d x) \\\\ \mathrm{E}_{\mathrm{PQ}} & =\frac{\mu_0 \mathrm{I} v}{2 \pi} \int_1^4 \frac{d x}{x} \\\\ \mathrm{E}_{\mathrm{PQ}} & =\frac{\mu_0 \mathrm{I} v}{2 \pi} \ln 4=\frac{\mu_0 \mathrm{I} v}{2 \pi} \times 2 \ln 2 \end{aligned} $

On substituting all the values in above, we get

$ \begin{aligned} \mathrm{E}_{\mathrm{PQ}} & =2 \times 10^{-7} \times 2 \times 3 \times 2 \times 0.7 \\\\ & =16.8 \times 10^{-7} \mathrm{~V} \end{aligned} $

At $t=0, \mathrm{C}$ acts as a short. So, maximum current flows through $R$.

$ \begin{aligned} i_{\max } & =\frac{\mathrm{E}_{\mathrm{PQ}}}{\mathrm{R}}=\frac{16.8 \times 10^{-7}}{1.4} \\\\ & =12 \times 10^{-7} \mathrm{~A} \\\\ & =1.2 \times 10^{-6} \mathrm{~A} \end{aligned} $

At $t \rightarrow \infty, C$ acts as on open. So,

$ \begin{aligned} q_{\max } & =\mathrm{CE}_{\mathrm{PQ}}=5 \times 10^{-6} \times 16.8 \times 10^{-7} \mathrm{C} \\\\ & =84 \times 10^{-13} \mathrm{C} \\\\ & =8.4 \times 10^{-12} \mathrm{C} \end{aligned} $

The magnetic field at the origin O (0, 0, 0) due to the infinitely long straight wire located at x = +R is

${\overrightarrow B _1}(O) = {{{\mu _0}{i_1}} \over {2\pi R}}\widehat k$

and due to the infinitely long straight wire located at x = $-$R is

${\overrightarrow B _2}(O) = {{{\mu _0}{i_2}} \over {2\pi R}}( - \widehat k)$.

The magnetic field at O due to the circular loop is

${\overrightarrow B _3}(O) = {{{\mu _0}i{R^2}} \over {2{{({R^2} + {{(\sqrt 3 R)}^2})}^{3/2}}}}( - \widehat k) = {{{\mu _0}i} \over {16R}}( - \widehat k)$.

The resultant magnetic field at the origin (0, 0, 0) is

${\overrightarrow B _3}(O) = {\overrightarrow B _1}(O) + {\overrightarrow B _2}(O) + {\overrightarrow B _3}(O)$

$ = {{{\mu _0}} \over {16\pi R}}(8({i_1} - {i_2}) - \pi i)\widehat k$.

Thus, if i₁ = i₂ then ${\overrightarrow B _1}(O) = - {\overrightarrow B _2}(O)$ and ${\overrightarrow B _1}(O) = {\overrightarrow B _3}(O) \ne \overrightarrow 0 $.

If i₁ > 0 and i₂ < 0 then both ${\overrightarrow B _1}(O)$ and ${\overrightarrow B _2}(O)$ are along $\widehat k$ but ${\overrightarrow B _3}(O)$ is along $-$$\widehat k$. Thus, resultant field $\overrightarrow B (O)$ can be zero.

If i₁ < 0 and i₂ > 0 then ${\overrightarrow B _1}(O)$, ${\overrightarrow B _2}(O)$ and ${\overrightarrow B _3}(O)$ are all along $-$$\widehat k$. Thus, resultant field $\overrightarrow B (O)$ cannot be zero. We encourage you to relate these results with the feasible solutions of 8(i₁ $-$ i₂) $-$ $\pi$i = 0 for i > 0.

Now, let us find the magnetic field at the centre of the loop i.e., point C in the figure.

The magnetic field $\overrightarrow B_1$ at the point C due to the infinitely long straight wire located x = +R is perpendicular to the line AC (see figure). It lies in the x-z plane and makes an angle 30$^\circ$ with the x axis. Resolve $\overrightarrow B_1$ along the x and z directions to get

${{\vec B}_1}(C) = {{{\mu _0}{i_1}} \over {2\pi (2R)}}(\cos 30^\circ \widehat i + \sin 30^\circ \widehat k)$.

Similarly, the magnetic field at C due to the infinitely long straight wire located at x = $-$R is

${{\vec B}_2}(C) = {{{\mu _0}{i_2}} \over {2\pi (2R)}}(\cos 30^\circ \widehat i - \sin 30^\circ \widehat k)$.

The magnetic field at C due to the circular loop is

${{\vec B}_3}(C) = {{{\mu _0}i} \over {(2R)}}( - \widehat k)$.

The resultant magnetic field at C is

${{\vec B}_1}(C) = {{\vec B}_1}(C) + {{\vec B}_2}(C) + {{\vec B}_3}(C) = {{{\mu _0}} \over {8\pi R}}\left( {\sqrt 3 ({i_1} + {i_2})\widehat i + ({i_1} - {i_2} - 4\pi i)\widehat k} \right)$.

If i₁ = i₂ then magnitudes of ${{\vec B}_1}(C)$ and ${{\vec B}_2}(C)$ are equal and their z-components are equal in magnitude but opposite in direction. Thus, z-component of the magnetic field at C is due to the circular loop only and its value is $\left( { - {{{\mu _0}i} \over {2R}}} \right)$.

As the magnetic field is perpendicular to motion of particle.

$\therefore$ Particle will move in curved path of radius,

$r = {{mv} \over {QB}} = {p \over {QB}}$ ($\because$ p = mv)

Particle will cross the region 2 if

$r > {{3R} \over 2} \Rightarrow B < {{2p} \over {3QR}}$

and particle will re-enter region 1 if

$r < {{3R} \over 2} \Rightarrow B > {{2p} \over {3QR}}$

Here a = 0, b = (r $-$ R)

Equation of trajectory, ${(x - a)^2} + {(y - b)^2} = {r^2}$

$\therefore$ To pass through region 2, at ${P_2}\left( {{{3R} \over 2},0} \right)$,

${\left( {{{3R} \over 2}} \right)^2} + {(r - R)^2} = {r^2}$ or ${9 \over 4}{R^2} + {r^2} + {R^2} - 2rR = {r^2}$

${{13} \over 4}{R^2} = 2rR$ or $r = {{13} \over 8}R \Rightarrow {p \over {QB}} = {{13} \over 8}R$ or $B = {8 \over {13}}{p \over {QR}}$

Hence, for $B = {8 \over {13}}{p \over {QR}}$ particle will enter region 3 through P₂

For $r < {{3R} \over 2}$, particle will re-enter region 1 through point P₃(x, y).

$y = (2r - R) = \left( {{{2mv} \over {QB}} - R} \right)$, x = 0

$\Rightarrow$ y $\propto$ m

At farthest point from y-axis, the momentum is perpendicular to initial momentum.

$\therefore$ $\Delta p = \sqrt 2 mv = \sqrt 2 p$

At equilibrium,

$ \begin{aligned} & \mathrm{F}_{\mathrm{M}} =\mathrm{F}_{\mathrm{E}} \\\\ & q \mathrm{v} \mathrm{B} =q \mathrm{E} \\\\ & \mathrm{E} =\mathrm{v} \cdot \mathrm{B} \\\\ & i =n e \mathrm{~A} v_d \end{aligned} $

where $v_d=$ drift speed of electrons and $n=$ electron density

$ \begin{aligned} \mathrm{E} & =\frac{i}{n e \mathrm{~A}} \mathrm{~B} \\ & =\frac{i}{n e(\omega d)} \end{aligned} $

Area of cross section is always taken normal to the direction of flow of current. Potential difference,

$\mathrm{V}=\mathrm{E} \omega=\frac{i \mathrm{~B}} { ned }$

For the two strips

$ \mathrm{V}_1=\frac{i \mathrm{~B}}{n e}\left(\frac{1}{d_1}\right)$

$\mathrm{V}_2=\frac{i B}{n e}\left(\frac{1}{d_2}\right)$

For option (A)

$ \frac{\mathrm{V}_1}{\mathrm{~V}_2}=\frac{d_2}{d_1}=\frac{1}{2} $

For option (B)

$ \frac{\mathrm{V}_1}{\mathrm{~V}_2}=1, \frac{d_2}{d_1}=\frac{1}{2} $

For option (C)

$ \frac{\mathrm{V}_1}{\mathrm{~V}_2}=\frac{1}{2}, \frac{d_2}{d_1}=1 $

For option (D)

$ \frac{\mathrm{V}_1}{\mathrm{~V}_2}=\frac{d_2}{d_1}=1 $

$ \text { and } \quad \begin{aligned} & \mathrm{V}_1 =\frac{i \mathrm{~B}_1}{n_1 e d} \\\\ & \mathrm{~V}_2 =\frac{i \mathrm{~B}_2}{n_2 e d} \\\\ & \frac{\mathrm{V}_1}{\mathrm{~V}_2} =\frac{\mathrm{B}_1}{\mathrm{~B}_2} \frac{n_2}{n_1} \end{aligned} $

For option (A) :

$ \frac{\mathrm{V}_1}{\mathrm{~V}_2}=\frac{1}{2} $

and $ \frac{\mathrm{B}_1}{\mathrm{~B}_2} \frac{n_2}{n_1}=1 \times \frac{1}{2}=\frac{1}{2}$

For option (B) :

$ \begin{aligned} \frac{\mathrm{V}_1}{\mathrm{~V}_2} & =1 \\\\ \text { and } \frac{\mathrm{B}_1}{\mathrm{~B}_2} \frac{n_2}{n_1} & =1 \times \frac{1}{2}=\frac{1}{2} \end{aligned} $

For option (C) :

$ \begin{aligned} \frac{\mathrm{V}_1}{\mathrm{~V}_2} & =2 \\\\ \text { and } \frac{\mathrm{B}_1}{\mathrm{~B}_2} \frac{n_2}{n_1} & =2 \times 1=2 \end{aligned} $

For option (D) :

$ \begin{aligned} \frac{\mathrm{V}_1}{\mathrm{~V}_2} & =1 \\\\ \text { and } \frac{\mathrm{B}_1}{\mathrm{~B}_2} \frac{n_2}{n_1} & =2 \times 1=2 \end{aligned} $

The force on a conducting element of length $d\overrightarrow l $, carrying a current I in a magnetic field $\overrightarrow B $, is given by d$\overrightarrow F $ = I d$\overrightarrow l $ $\times$ $\overrightarrow B $. If the field is uniform, the total force on the conductor is given by

$\overrightarrow F = \int_a^g {Id\overrightarrow l \times \overrightarrow B = I\left( {\int_a^g {d\overrightarrow l } } \right) \times \overrightarrow B = I\,\overrightarrow {ag} \times \overrightarrow B } $

Note that $\overrightarrow B $ is taken out of the integral sign because it is constant. The vector $\overrightarrow {ag} = 2(L + R)\widehat x$. The magnetic forces on the conductor in the given cases are

Case (A) : $\overrightarrow F = I(2(L + R)\widehat x) \times (B\widehat z)$

$ = - 2IB(L + R)\widehat y$

Case (B) : $\overrightarrow F = I(2(L + R)\widehat x) \times (B\widehat x) = \overrightarrow 0 $

Case (C) : $\overrightarrow F = I(2(L + R)\widehat x) \times (B\widehat y)$

$ = 2IB(L + R)\widehat z$

We know that

(i) For a cylinder, B = 0 for 0 < r < R and $B = {{{\mu _0}} \over {4\pi }} \times {{2I} \over r}$ for r > R.

(ii) For a solenoid, B = $\mu$₀nI for 0 < r < R and B = 0 for r > R.

$\bullet$ Assuming that the magnetic induction to be B_C for cylinder and B_S for solenoid.

In the region R > r > 0, B_S = $\mu$₀nI while B_C = 0. Therefore, the total field is non-zero.

Hence, option (A) is correct.

$\bullet$ In the region 2R < r < R, the magnetic field is not along the axis of cylinder which is given as

$B = \sqrt {B_S^2 + B_C^2} $

Hence, option (B) is wrong.

$\bullet$ For checking option (C), in the region R < r < 2R, the magnetic field is tangential to the circle of radius r, centred on the axis. So this statement is wrong because magnetic field is not in the plane of circle.

Hence, option (C) is wrong.

$\bullet$ In the region r > 2, B_S = 0; while ${B_C} = {{{\mu _0}} \over {4\pi }} \times {{2I} \over r}$. Therefore total field is non-zero.

Hence, option (D) is correct.

The situation is as shown in the figure.

Here, ${\overrightarrow u _1} = 4\widehat i$ ms^$-$1

${\overrightarrow u _2} = 2\left( {\sqrt 3 \widehat i + \widehat j} \right)$ ms^$-$1

When the charged particle enters in an uniform magnetic field, it travels a circular path.

Component of final velocity of particle is in positive y direction.

Centre of circle is present on positive y-axis.

So, magnetic field is present in negative z-direction.

Angle of deviation

$\tan \theta = {{{u_{2y}}} \over {{u_{2x}}}} = {2 \over {2\sqrt 3 }} = {1 \over {\sqrt 3 }}$

$\theta = {\tan ^{ - 1}}\left( {{1 \over {\sqrt 3 }}} \right) = 30^\circ = {\pi \over 6}$

$\omega = {\theta \over t}$

$\therefore$ $t = {\theta \over \omega } = {\theta \over {(QB/M)}} = {{M\theta } \over {QB}}$ ($\because$ $\omega = {{QB} \over M}$

$\therefore$ $B = {{M\theta } \over {Qt}} = {{M\pi } \over {6Q \times 10 \times {{10}^{ - 3}}}}$ (.... t = 10 ms = 10^$-$3 s)

$ = {{100\pi M} \over {6Q}} = {{50\pi M} \over {3Q}}$

For $\theta$ = 90$^\circ$ : $\overrightarrow v \,||\,\overrightarrow E $ and $\overrightarrow v \,||\,\overrightarrow B $

. Therefore, magnetic field $\overrightarrow B $ exerts no force on the point charge :

${\overrightarrow F _B} = q(\overrightarrow v \times \overrightarrow B ) = 0$

${\overrightarrow F _E} = q\overrightarrow E $

Therefore, the motion is along y-axis and the particle accelerates.

For $\theta$ = 0$^\circ$ and $\theta$ = 10$^\circ$ : The motion is helical with increasing pitch, along with increasing time (along y-axis), as $p = (v\cos \theta )t$, where v increases due to ${\overrightarrow F _E}$. (For $\theta$ = 0$^\circ$, the change cannot move in a circular path.)

$r = {{mv} \over {Bq}}$

or, $r \propto m$

$\therefore$ ${r_e} < {r_p}$ as ${m_e} < {m_p}$

Further, $T = {{2\pi m} \over {Bq}}$

or $T \propto m$

$\therefore$ ${T_e} < {T_p}$

${t_e} = {{{T_e}} \over 2}$

and ${t_p} = {{{T_p}} \over 2}$

or ${t_e} < {t_p}$

$\therefore$ Correct options are (b) and (d).

As the particle enters the magnetic field, a force acts on it due to the magnetic field which moves the particle in a circular path of radius.

$r={{mv} \over {qB}}$

For the particle entering in region III, r > $l$

$ \Rightarrow {{mv} \over {qB}} > l$

$ \Rightarrow v > qlB/m$

For maximum path length in region II, r = $l$

$\therefore$ $l = {{mv} \over {qB}}$

$ \Rightarrow v = {{qlB} \over m}$

[$\because$ ${v_{\max }} = {{qBl} \over m},T = {{2\pi r} \over v} = {{2\pi m} \over {q.B}}$]

The time taken by the particle to move in region II before coming back in region I is given by $t=\pi m$, which is independent of v.

From right hand rule, we write, the force on a current carrying wire placed in a magnetic field is given by $\overrightarrow{\mathrm{F}}=(l \times \mathrm{B}) \hat{i}$

If $\phi$ is angle between $l$ and B , then $\overrightarrow{\mathrm{F}}=i l \mathrm{~B} \sin \phi$ Force on $\mathrm{AB}=\mathrm{BIL}=\frac{\mu_0 \mathrm{I}_1}{2 \pi r_1} \mathrm{IL}_{\mathrm{AB}}$.

$ \mathrm{F}_{\mathrm{AB}}=\frac{\mu_0 \mathrm{I}_1}{2 \pi r_1} \mathrm{I}\left(\frac{\phi}{2 \pi} 2 \pi r_1\right)=\frac{\mu_0 \mathrm{I} \mathrm{I}_1 \phi}{2 \pi r} $

(This is inward force)

Force on $\mathrm{CD}=\mathrm{BIL}=\frac{\mu_0 \mathrm{I}_1}{2 \pi r_2} \mathrm{IL}_{\mathrm{CD}}$.

$ \mathrm{F}_{\mathrm{CD}}=\frac{\mu_0 \mathrm{I}_1}{2 \pi r_2}\left(\frac{\phi}{2 \pi} 2 \pi r_2\right)=\frac{\mu_0 \mathrm{II}_1 \phi}{2 \pi} $

(This is out wards)

Here, net force is zero.

$B C$ and $C D$ are radius vectors. The $I$ and $B$ are $\perp$ er here,

$ \therefore=i l \mathrm{~B} \sin 90^{\circ}=i l \mathrm{~B} . $

The force acting on the both section is same as the $y$ experience same magnetic field at every point on the wire.

But the direction is different for both of them.

$\therefore$ Force on BC acts upwards and AD acts downwards.

Thus it appears rotating clockwise as scene from O .