NEET Practice Q & A

Hint

(a) For hollow metallic cylinder, magnetic field inside is zero while outside it the magnetic field is inversely proportional to distance from centre of cylinder.
So, variation is correctly shown by graph (a).

Hint

(a) outside the cable. In a coaxial cable, the central conductor and the outer conductor carry equal currents in opposite directions. The magnetic field outside the cable is zero because the magnetic fields generated by the two conductors cancel each other out due to the opposite directions of the currents.

Hint

\(
\begin{aligned}
& \text { (c) We have, } B_1=\frac{\mu_0 I}{2 r}, B_2=\frac{\mu_0 I}{4 r} \\
& \therefore \quad \frac{B_1}{B_2}=2
\end{aligned}
\)

Hint

(a) Magnetic force on wire \(B\) from both the wires is towards \(A\).

Hint

(d) Given, \(2 \pi R=2.0 \mathrm{~m}\)
\(
\therefore \quad R=\frac{1.0}{\pi} \mathrm{~m}
\)
Magnetic moment, \(M=i A=(i)\left(\pi R^2\right)\)
\(
=(1)(\pi)\left(\frac{1.0}{\pi}\right)^2=\frac{1}{\pi} \mathrm{~A}-\mathrm{m}^2
\)

Hint

(c) Magnetic field produced at the centre of the orbit,
\(
B=\frac{\mu_0}{4 \pi} \frac{2 \pi i}{r}
\)
Now, \(i=\frac{q}{t}=q f\)
\(
\begin{aligned}
\therefore \quad B & =\frac{\mu_0}{4 \pi} \frac{2 \pi(q f)}{r} \\
& =\frac{4 \pi \times 10^{-7} \times 2 \times 3.14 \times 1.6 \times 10^{-19} \times 6.6 \times 10^{15}}{4 \pi \times 0.53 \times 10^{-10}} \\
& =12.5 \mathrm{~Wb} \mathrm{~m}^{-2}
\end{aligned}
\)

Hint

(b) Force or tension in the loop, \(F=\mathrm{Bil}\)
\(
\Rightarrow \quad F \propto B
\)
If \(B\) is doubled, then tension in the loop is double.

Hint

(a) In Ist case
\(
l=2 \pi r_1 \dots(i)
\)
At centre, \(\quad B=\frac{\mu_0 i}{2 r_1}\)
In IInd case
\(
l=2 \times \pi r_2 \dots(ii)
\)
From Eqs. (i) and (ii), we get
\(
\begin{aligned}
& & 2 \times 2 \pi r_2 & =2 \pi r_1 \Rightarrow r_2=r_{1 / 2} \\
& \therefore & & B^{\prime}=\frac{2 \times \mu_0 i}{2 r_2} \text { ( } \because \text { there are } 2 \text { turns) } \\
& \Rightarrow & & \frac{B^{\prime}}{B}=\frac{2 r_1}{r_2}=4 \Rightarrow B^{\prime}=4 B
\end{aligned}
\)

Hint

(b) When wire is turned into \(n\) circular loops, then magnetic field produced is \(B^{\prime}=n^2 B=(3)^2 B_1=9 B_1\).

Hint

(c) As magnetic field and velocity of charge is in same direction, so magnetic force on it is zero. Hence, it falls with acceleration equal to \(g\).

Hint

(d) Velocity vector is parallel to magnetic field vector.
\(
\therefore \quad \theta=0^{\circ}
\)
Hence, force, \(F=B q v \sin 0=0\)

Hint

(b) In this case, the particle moves with velocity \(\overrightarrow{\boldsymbol{u}}\) along the positive \(x\)-direction ( \(u_x=u_1 u_y=0, u_z=0\) ) and the magnetic field is along the negative \(z\)-direction ( \(\overrightarrow{\boldsymbol{B}}=-\boldsymbol{B} \hat{\mathbf{z}}\) ), where B is the magnitude of the magnetic field.
The Lorentz force is then:
\(
\begin{aligned}
& \vec{F}=q(\vec{u} \times \vec{B})=q(u \hat{x} \times(-B \hat{z})) \\
& \vec{F}=q(-u B(\hat{x} \times \hat{z}))
\end{aligned}
\)
Using the right-hand rule for the cross product, \(\hat{x} \times \hat{z}=-\hat{y}\).
\(
\vec{F}=q(-u B(-\hat{y}))=q u B \hat{y}
\)
The force is in the positive \(y\)-direction. This force provides the centripetal acceleration for the circular motion.
The magnitude of the magnetic force is \(\boldsymbol{F}=|\boldsymbol{q}| \boldsymbol{u} \boldsymbol{B}\) (since the velocity is perpendicular to the magnetic field, the \(\boldsymbol{\operatorname { s i n }} \boldsymbol{\theta}\) term is 1 ). This force is equal to the centripetal force, \(\boldsymbol{F}_c=\frac{\boldsymbol{m} \boldsymbol{u}^2}{\boldsymbol{R}}\).
Therefore, \(|q| \boldsymbol{u B}=\frac{\boldsymbol{m} \boldsymbol{u}^2}{\boldsymbol{R}}\).
Solving for the radius of the circular path, \(R=\frac{m \boldsymbol{u}}{|q| \boldsymbol{B}}\).
The particle enters the magnetic field region at \(\boldsymbol{x}=\boldsymbol{a}\) and moves along the positive x -direction. For the particle to just enter the region \(\boldsymbol{x} \boldsymbol{>} \boldsymbol{b}\), the radius of the circular path must be equal to the width of the magnetic field region, which is \(\boldsymbol{b}-\boldsymbol{a}\).
So, \(\boldsymbol{R}=\boldsymbol{b}-\boldsymbol{a}\).
Substituting the expression for the radius: \(\frac{\boldsymbol{m} \boldsymbol{u}}{|\boldsymbol{q}| \boldsymbol{B}}=\boldsymbol{b}-\boldsymbol{a}\).
Solving for \(u: u=\frac{(b-a)|q| B}{m}\).

Hint

(c) Radius of circular path,
\(
\begin{aligned}
r & =\frac{m v \cos 60^{\circ}}{B q} \\
& =\frac{\left(1.67 \times 10^{-27}\right)\left(2 \times 10^6\right) \cos 60^{\circ}}{0.14 \times 1.6 \times 10^{-19}}=0.07 \mathrm{~m} \\
T & =\frac{2 \pi m}{B q}=\frac{(2 \pi)\left(1.67 \times 10^{-27}\right)}{0.14 \times 1.6 \times 10^{-19}}=0.5 \times 10^{-6} \mathrm{~s}
\end{aligned}
\)

Hint

(d) Distribution of current will be as shown below.

\(
\begin{aligned}
&\text { Net magnitude of magnetic field at the centre } O \text { is given by }\\
&\begin{aligned}
B_{\text {het }} & =B_{a b}+B_{l c}-B_{a c} \\
& =\frac{\mu_0}{2 \pi} \cdot \frac{I / 3}{C O}+\frac{\mu_0}{2 \pi} \cdot \frac{I / 3}{A O}-\frac{\mu_0}{2 \pi} \cdot \frac{2 I / 3}{B O} \\
& =\frac{\mu_0}{2 \pi} \cdot \frac{I / 3}{r}+\frac{\mu_0}{2 \pi} \cdot \frac{I / 3}{r}-\frac{\mu_0}{2 \pi} \cdot \frac{2 I / 3}{r} [\because A O=B O=C O=r] \\
& =0
\end{aligned}
\end{aligned}
\)

Hint

(a)

\(
\begin{aligned}
B & =(\text { due to circular wire })+(\text { due to straight wire }) \\
& =\frac{\mu_0 I}{2 R}+\frac{\mu_0}{2 \pi} \frac{I}{R}=\frac{\mu_0 I}{2 R}\left(1+\frac{1}{\pi}\right)=\frac{4 \mu_0 I}{4 R \pi}(1+\pi)
\end{aligned}
\)

Hint

(d) The magnitude of the resultant magentic field at the common centre of coil is given as \(B_{\text {het }}=\sqrt{2} B\)

The ratio of the magnitude of the resultant magnetic field at the common centre and the magnetic field due to one coil alone at common centre.
\(
\Rightarrow \quad \frac{B_{\text {net }}}{B_{\text {at centre }}}=\frac{\sqrt{2} B}{B}=\frac{\sqrt{2}}{1}
\)

Hint

\(
\text { (c) Magnetic field, } B=\frac{\mu_0 i}{2 R}=\frac{\mu_0(e f)}{2 R}=\frac{\left(\mu_0\right)(e)\left(\frac{v}{2 \pi R}\right)}{R}
\)
\(
\begin{array}{ll}
\Rightarrow & R^2=\frac{\mu e v}{2 \pi B} \\
\Rightarrow & R^2 \propto \frac{v}{B} \\
\Rightarrow & R \propto \sqrt{\frac{v}{B}}
\end{array}
\)

Hint

(c) Let the length of each wire be \(L\).
For square, length of each side \(=L / 4\)
Area of square \(=\left(\frac{L}{4}\right)^2=\frac{L^2}{16}\)
For circle, \(L=2 \pi r \Rightarrow r=\frac{L}{2 \pi}\)
Area of circle \(=\pi r^2=\pi\left(\frac{L}{2 \pi}\right)^2=\frac{L^2}{4 \pi}\)
As, magnetic moment, \(M=i A\)
\(
\begin{array}{ll}
\Rightarrow & M \propto A \\
\therefore & \frac{M_{\text {square }}}{M_{\text {circle }}}=\frac{A_{\text {square }}}{A_{\text {circle }}}=\frac{L^2 / 16}{L^2 / 4 \pi}=\frac{\pi}{4}
\end{array}
\)

Hint

(b) At \(O\), due to wire \(A B\) and \(D E\), magnetic field will be zero. The combined effect of \(B C\) and \(D F\) is equivalent to that of an infinitely long wire, i.e. both fields are in same direction.
\(
\Rightarrow \quad B_O=\frac{2 \mu_0 I}{4 \pi r}=\frac{\mu_0 I}{2 \pi r}
\)

Hint

(d) The given figure is as shown

\(
\frac{R_1}{R_2}=\frac{1}{3} \quad[\because \text { Resistance, } R \propto l]
\)
\(
\frac{I_1}{I_2}=\frac{R_2}{R_1}=3 \quad[\because V=4 \mathrm{~V}]
\)
\(
\begin{array}{ll}
\therefore & B_1=\frac{\mu_0 I_1}{2 r} \times \frac{90^{\circ}}{360} \otimes \text { and } B_2=\frac{\mu_0 I_2}{2 r} \times \frac{270}{360} . \\
\therefore & \frac{B_1}{B_2}=\frac{I_1}{I_2} \times \frac{90}{270}=3 \times \frac{1}{3}=1
\end{array}
\)
or \(\quad B_1 \otimes=B_2 \odot\)
\(\therefore\) Net magnetic field at centre of conductor is zero.

Hint

\(
\text { (b) } F_a=F_{a b}-F_{a c}=\frac{\mu_0 I^2 l}{2 \pi d}-\frac{\mu_0 I^2 l}{2 \pi(2 d)}=\frac{\mu_0 I^2 l}{4 \pi d}
\)

\(
F_b=F_{b a}+F_{b c}=\frac{\mu_0 I^2 l}{2 \pi d}+\frac{\mu_0 I^2 l}{2 \pi d}=\frac{\mu_0 I^2 l}{\pi d}
\)
\(
\begin{aligned}
&F_c=F_{c a}+F_{c b}=\frac{\mu_0 I^2 l}{2 \pi(2 d)}+\frac{\mu_0 I^2 l}{2 \pi d}=\frac{3 \mu_0 I^2 l}{4 \pi d}\\
&\text { On comparing values as } F_a, F_b \text { and } F_c \text {, we get }\\
&F_b>F_c>F_a
\end{aligned}
\)

Hint

(c) \(F_1=\frac{\mu_0}{4 \pi} \frac{I^2 \times l}{x}\), towards \(A\) and \(F_2=\frac{\mu_0}{4 \pi} \frac{2 I^2 \times l}{2 x}\), towards \(A\)
\(\therefore \quad F_1=F_2\)

Hint

(c)

Force on wire \(Q\) due to wire \(R\),
\(
\begin{aligned}
F_R & =\frac{4 \pi \times 10^{-7}}{4 \pi} \times \frac{2 \times 20 \times 10}{0.02} \times 0.1 \\
& =20 \times 10^{-5} \mathrm{~N} \quad \text { (towards right) }
\end{aligned}
\)
Force on wire \(Q\) due to wire \(P\),
\(
\begin{aligned}
F_P & =\frac{4 \pi \times 10^{-7}}{4 \pi} \times \frac{2 \times 30 \times 10}{0.1} \times 0.1 \\
& =6 \times 10^{-5} \mathrm{~N} \text { (towards right) }
\end{aligned}
\)
Net force on \(Q\),
\(
F=F_R+F_P=20 \times 10^{-5}+6 \times 10^{-5}=26 \times 10^{-5} \mathrm{~N}
\)
\(
=2.6 \times 10^{-4} \mathrm{~N} \quad \text { (towards right) }
\)

Hint

(d) If charge \(q\) is positive, then due to electric field their velocity will be towards positive \(X\)-axis.
Now, \(\mathbf{F}_m=q(\mathbf{v} \times \mathbf{B})\)
where, \(\quad \mathbf{v} \rightarrow\) towards \(\hat{\mathbf{i}}\)
and \(\quad \mathbf{B} \rightarrow\) towards \(\hat{\mathbf{k}}\)
\(
\therefore \quad \mathbf{F}_m \rightarrow \text { towards }-\hat{\mathbf{j}} \text { (i.e. }-y \text {-direction) }
\)
Similarly, we can see with – \(q\) charge.

Hint

(c) The motion of a charged particle in a magnetic field is helical.
The pitch of the helix is the distance traveled along the magnetic field direction in one revolution.
The pitch is given by \(p=\frac{2 \pi m v \cos (\theta)}{q B}\), where \(\theta\) is the angle between the velocity and the magnetic field.
For minimum value of \(B\), it is colliding after one pitch. It will collide the target again, if pitch is halved or remains same.
Find the pitch of the helix
The pitch is given by \(p=\frac{2 \pi m v \cos (\theta)}{q B}\).
Let \(d\) be the distance to the target.
Then \(d=\frac{2 \pi m v_0 \cos (\theta)}{q B_0}\).
\(
p \propto \frac{v}{B}
\)
For option (a), \(\boldsymbol{B}=2 \boldsymbol{B}_0\) and \(v=2 v_0\). The pitch is \(p_a=\frac{2 \pi m\left(2 v_0\right) \cos (\theta)}{q\left(2 B_0\right)}=\frac{2 \pi m v_0 \cos (\theta)}{q B_0}=d\). Thus, the particle will strike the target.
For option (b), \(\boldsymbol{B}=2 \boldsymbol{B}_0\) and \(v=v_0\).
The pitch is \(p_b=\frac{2 \pi m v_0 \cos (\theta)}{q\left(2 B_0\right)}=\frac{1}{2} \frac{2 \pi m v_0 \cos (\theta)}{q B_0}=\frac{1}{2} d\).
\(
\text { For } p / p_0=1 \text { and } \frac{1}{2} \text {, both options (a) and (b) are correct. }
\)

Hint

(b) \(\tau=\mathbf{M} \times \mathbf{B}=I A B\)
Magnetic moment is in positive \(x\)-direction and the magnetic field in positive \(y\)-direction. So, the torque (IAB) must be in positive \(z\)-direction.

Hint

\(
\begin{aligned}
& \text { (c) We know that, current, } i=q f \\
& \quad=1.6 \times 10^{-19} \times 6.6 \times 10^{15}=10.5 \times 10^{-4} \mathrm{~A} \\
& \therefore \text { Area, } A=\pi R^2=3.14 \times(0.528)^2 \times 10^{-20} \mathrm{~m}^2
\end{aligned}
\)
\(\therefore\) Magnetic moment, \(M=i A\)
\(
\begin{aligned}
& =10.5 \times 10^{-4} \times 3.14 \times(0.528)^2 \times 10^{-20} \\
& =10 \times 10^{-24} \text { units }=1 \times 10^{-23} \text { units }
\end{aligned}
\)

Hint

(a) All magnetic fields are cancelling each other.

Hint

\(
\text { (d) Here, diameter of circular path, } y=2 r=2 \frac{m v_0}{q B_0}
\)

Hint

(a) Torque acting on the coil, \(\tau=NiBA\)
\(
=100 \times 2 \times 0.2 \times(0.08 \times 0.1)=0.32 \mathrm{~N}-\mathrm{m}
\)
According to Fleming’s left hand rule, force on side \(A D\) is in upward direction out of the page and force on side \(B C\) is in downward direction into the page.
Hence, this torque rotates the side \(A D\) out of the page.

Hint

\(
\text { (b) Given, time, } t=\frac{\pi m}{B q}=\frac{T}{2} \text { of proton }
\)

\(
\text { Sepearation, } d=\sqrt{2} r=\frac{\sqrt{2} m v}{B q}
\)

Hint

\(
\begin{aligned}
& \text { (b) As, } d A=(2 \pi r) d r \\
& \therefore \quad d q=\sigma \cdot d A=\left(2 \pi \alpha r^3\right) \cdot d r \\
& \text { Current, } i=(d q) f=(d q) \frac{\omega}{2 \pi}=\left(\alpha \omega r^3\right) d r \\
& \text { Now, } d B=\frac{\mu_0 i}{2 r}=\frac{\alpha \mu_0 \omega r^2 d r}{2} \\
& \therefore \quad B=\int_0^R d B=\frac{\mu_0 \alpha \omega R^3}{6}
\end{aligned}
\)

Hint

(b)

\(
\begin{aligned}
&\begin{aligned}
\vec{M} & =i\left(\pi r^2\right) \hat{k} \\
\vec{B} & =B_x \hat{i}+B_z \hat{k}
\end{aligned}\\
&\text { To start tilting }
\end{aligned}
\)
\(
\begin{aligned}
& \tau_{\text {due to mass }}=\tau_{\text {magnetic }} \\
& |(m g) r|=|\vec{M} \times \vec{B}| \\
& (m g)(r)=i\left(\pi r^2\right)\left(B_x\right) \\
& i=\frac{m g}{\pi r B_x}
\end{aligned}
\)

Hint

(b) At, the centre of coil 1,
\(
B_1=\frac{\mu_0}{4 \pi} \times \frac{2 \pi i_1}{r_1} \dots(i)
\)
At the centre of coil-2,
\(
B_2=\frac{\mu_0}{4 \pi} \times \frac{2 \pi i_2}{r_2} \dots(ii)
\)
But \(B_1=B_2\)
\(
\therefore \quad \frac{\mu_0}{4 \pi} \frac{2 \pi i_1}{r_1}=\frac{\mu_0}{4 \pi} \frac{2 \pi i_2}{r_2} \text { or } \frac{i_1}{r_1}=\frac{i_2}{r_2}
\)
\(
\begin{array}{ll}
\text { As, } & r_1=2 r_2 \\
\therefore & \frac{i_1}{2 r_2}=\frac{i_2}{r_2} \text { or } i_1=2 i_2
\end{array}
\)
\(
\begin{aligned}
&\text { Now, ratio of potential differences, }\\
&\frac{V_2}{V_1}=\frac{i_2 \times R_2}{i_1 \times R_1}=\frac{i_2 \times R_2}{2 i_2 \times 2 R_2}=\frac{1}{4} [\because \text { Resistance, } R \propto r \text { (radius) }]
\end{aligned}
\)
\(
\Rightarrow \quad \frac{V_1}{V_2}=\frac{4}{1}
\)

Hint

\(
\begin{aligned}
&\text { (d) Velocity, } v=\frac{E}{B} \text { and radius of path, }\\
&r=\frac{m v}{B q}=\frac{v}{B \cdot q / m}=\frac{E / B}{B \cdot S}=\frac{E}{B^2 S}
\end{aligned}
\)

Hint

(b) Applying Ampere’s circuital law in the closed loop as shown in figure,

\(
\begin{aligned}
& \qquad 2 B l=\mu_0(\lambda \cdot l) \\
& \therefore \text { Magnetic field induction, } B=\frac{\mu_0 \lambda}{2}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \text { (a) We have, } \mathbf{F}_{C A D}=\mathbf{F}_{C D}=\mathbf{F}_{C E D} \\
& \therefore \text { Net force on frame }=3 \mathbf{F}_{C D} \\
& \quad=3(\text { Bil })=3 \times 4 \times 2 \times 1=24 \mathrm{~N}
\end{aligned}
\)

Hint

(c) According to the question,

\(
m g \sin \theta=I l B \cos \theta
\)
\(
\begin{aligned}
&\therefore \text { Magnitude of magnetic field, } B=\frac{m g}{I l} \tan \theta\\
&=\frac{m g \tan 60^{\circ}}{2 \times(2 L)}=\frac{\sqrt{3} m g}{4 L}
\end{aligned}
\)

Hint

(a) Let the magnetic field,
\(
\mathbf{B}=B_1 \hat{\mathbf{i}}+B_2 \hat{\mathbf{j}}+B_3 \hat{\mathbf{k}}
\)
Applying \(\mathbf{F}_m=q(\mathbf{v} \times \mathbf{B})\) two times, we have
\(
q[-\hat{\mathbf{j}}+\hat{\mathbf{k}}]=q\left[(\hat{\mathbf{i}}) \times\left(B_1 \hat{\mathbf{i}}+B_2 \hat{\mathbf{j}}+B_3 \hat{\mathbf{k}}\right)\right]=q\left[B_2 \hat{\mathbf{k}}-B_3 \hat{\mathbf{j}}\right]
\)
\(
\begin{aligned}
&\text { On comparing two sides, we get }\\
&\begin{aligned}
B_2 & =1 \text { and } B_3=1 \\
\text { Further, } q[\hat{\mathbf{i}}-\hat{\mathbf{k}}] & =q\left[(\hat{\mathbf{j}}) \times\left(B_1 \hat{\mathbf{i}}+B_2 \hat{\mathbf{j}}+B_3 \hat{\mathbf{k}}\right)\right] \\
& =q\left[-B_1 \hat{\mathbf{k}}+B_3 \hat{\mathbf{i}}\right]
\end{aligned}
\end{aligned}
\)
\(
\begin{aligned}
&\text { Again comparing, we get }\\
&B_1=1 \text { and } B_3=1
\end{aligned}
\)
\(
\therefore \quad \mathrm{B}=(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}) \mathrm{Wb} / \mathrm{m}^2
\)

Hint

\(
\text { (b) According to question, } l=2 \pi R \Rightarrow R=\frac{l}{2 \pi}
\)

\(
\begin{aligned}
&\text { and }\\
&\begin{aligned}
l & =4 a \Rightarrow a=\frac{l}{4} \\
B^{\prime} & =\frac{\mu_0 i}{2 R}=\frac{\mu_0 i}{2 \cdot \frac{l}{2 \pi}}=\frac{\mu_0 \pi i}{l} \\
B & =4\left[\frac{\mu_0}{4 \pi} \cdot \frac{i}{a / 2}\left(\sin 45^{\circ}+\sin 45^{\circ}\right)\right] \\
& =\frac{2 \sqrt{2} \mu_0 i}{\pi a}=\frac{8 \sqrt{2} \mu_0 i}{\pi l} \left[\because a=\frac{l}{4}\right]
\end{aligned}
\end{aligned}
\)
\(
\therefore \quad \frac{B}{B^{\prime}}=\frac{8 \sqrt{2}}{\pi^2}
\)

Hint

\(
\text { (c) Here, } \mathbf{F}_{b c}+\mathbf{F}_{d a}=0
\)

Force, \(\mathbf{F}_{a b}=I l B_0\left(1+\frac{x}{l}\right)=I B_0 l+I B_0 x \quad\) (towards right)
Force, \(\mathbf{F}_{c d}=I l B_0\left[1+\frac{x+l}{l}\right]=2 I B_0 l+I B_0 x \quad\) (towards left)
\(\therefore \quad \mathbf{F}_{\text {net }}=\mathbf{F}_{c d}-\mathbf{F}_{a b}=I B_0 l \quad\) (towards left)

Hint

(d) At centre, \(B=0\).
As, \(x \rightarrow a, B \rightarrow 0\) between centre and \(x=a\), there will be a maximum value.
Along \(X\)-axis for \(x>0, B_{\text {het }}\) is downwards.
Along \(X\)-axis for \(x<0, B_{\text {net }}\) is upwards.

Hint

\(
\text { (b) Radius, } r=a \cos \frac{\pi}{n}
\)

\(
\begin{aligned}
&\therefore \text { Magnetic field, }\\
&\begin{aligned}
B & =n\left[\frac{\mu_0}{4 \pi} \cdot \frac{i}{a \cos \pi / n}\left(\sin \frac{\pi}{n}+\sin \frac{\pi}{n}\right)\right] \\
& =\frac{\mu_0 n i}{2 \pi a} \tan \left(\frac{\pi}{n}\right)
\end{aligned}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (c) The magnitude of the force acting on the wire, }\\
&F_{\mathrm{net}}=\sqrt{F^2+F^2}=\sqrt{2} F=\sqrt{2} I l B
\end{aligned}
\)

Hint

\(
\text { (b) Magnetic field at } O \text {, }
\)

\(
\begin{aligned}
B & =2\left[\begin{array}{r}
\frac{\mu_0}{4 \pi} \frac{2 I / 3}{a / 2}\left(\sin 45^{\circ}+\sin 45^{\circ}\right)-\frac{\mu_0}{4 \pi} \cdot \frac{I / 3}{a / 2} \\
\left(\sin 45^{\circ}+\sin 45^{\circ}\right)
\end{array}\right] \\
& =\frac{\sqrt{2} \mu_0 I}{3 \pi a}
\end{aligned}
\)

Hint

(b) As, \(k x_0=m g=I L B=\left(\frac{E}{R}\right) \cdot L B\)
\(\therefore\) The magnetic field, \(B=\frac{m g R}{E L}\)

Hint

(a) Motion of charged particle in magnetic field \((\mathbf{B}=-2 \hat{\mathbf{k}} \mathrm{~T})\) is shown below

Time period, \(T=\frac{2 \pi m}{B q}=\frac{2 \pi}{B \cdot \frac{q}{m}}=\frac{2 \pi}{2 \cdot \pi}=1 \mathrm{~s}\)
Since, the particle will be at point \(P\) after time, \(t=\frac{1}{6} \mathrm{~s}=\frac{T}{6} \mathrm{~s}\)
Hence, if it is deviated by angle \(\theta\), then \(\theta=\frac{2 \pi}{6}=60^{\circ}\).
Therefore, velocity of the particle after \(t=\frac{1}{6} \mathrm{~s}\), is given by
\(
\begin{aligned}
\mathbf{v} & =10\left(\cos 60^{\circ} \hat{\mathbf{i}}+\sin 60^{\circ} \hat{\mathbf{j}}\right) \\
& =10\left(\frac{\hat{\mathbf{i}}}{2}+\frac{\hat{\mathbf{j}} \sqrt{3}}{2}\right) \\
& =(5 \hat{\mathbf{i}}+5 \sqrt{3} \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}
\end{aligned}
\)

Hint

\(
\text { (d) Deviation, } \theta=\sin ^{-1}\left(\frac{x}{R}\right)=\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)=60^{\circ}
\)

Velocity of the particle when it comes out of the field,

\(
\begin{aligned}
|\Delta \mathbf{v}|=2 v_0 & \sin \frac{\theta}{2} \\
& =2 v_0 \sin 30^{\circ}=v_0
\end{aligned}
\)

Hint

(c) As proton is moving with constant velocity, so acceleration is zero.
When \(E \neq 0, B=0\), then proton will move with a acceleration. Hence, this region of space for the motion proton in given condition does not satisity.
So, \(\quad E=0\) and \(B=0\)
When \(E=0\) and \(B \neq 0\) but proton is moving in parallel to the direction of magnetic field, then there will be no net force acting on proton.
When, \(E \neq 0\) and \(B \neq 0\) but electric force and magnetic force cancel each other, then also velocity will remain unchanged.

Hint

(d) We can consider the non-planar loop as made by two loops \(A B C D A\) and \(A Q R B A\) as shown in the figure.

Hint

(a)

\(
\begin{aligned}
&\text { (a) Area vector, }\\
&\begin{aligned}
\mathbf{A} & =\mathbf{C O} \times \mathbf{O A} \\
& =\left(-b \cos 37^{\circ} \hat{\mathbf{i}}-b \sin 37^{\circ} \hat{\mathbf{k}}\right) \times(a \hat{\mathbf{j}}) \\
& =-\frac{a b}{5}[4(\hat{\mathbf{i}} \times \hat{\mathbf{j}})+3(\hat{\mathbf{k}} \times \hat{\mathbf{j}})] \left[\because \cos 37^{\circ}=\frac{4}{5} \text { and } \sin 37^{\circ}=\frac{3}{5}\right]
\end{aligned}
\end{aligned}
\)
\(
\begin{aligned}
& =-\frac{a b}{5}[4 \hat{\mathbf{k}}-3 \hat{\mathbf{i}}] \\
& =\frac{a b}{5}[3 \hat{\mathbf{i}}-4 \hat{\mathbf{k}}]
\end{aligned}
\)
Magnetic moment, \(\mathbf{M}=N i_0 \mathbf{A}\)
\(
\begin{aligned}
& =\frac{N i_0 a b}{5}(3 \hat{\mathbf{i}}-4 \hat{\mathbf{k}}) \\
\mathbf{B} & =B_0 \hat{\mathbf{i}}
\end{aligned}
\)
\(\therefore\) Torque,
\(
\begin{aligned}
\tau & =\mathbf{M} \times \mathbf{B}=\frac{N i_0 a b}{5}(3 \hat{\mathbf{i}}-4 \hat{\mathbf{k}}) \times B_0 \hat{\mathbf{i}} \\
& =\frac{N i_0 a b B_0}{5}[3(\hat{\mathbf{i}} \times \hat{\mathbf{i}})-4(\hat{\mathbf{k}} \times \hat{\mathbf{i}})] \\
& =-\frac{4 N i_0 a b B_0}{5} \hat{\mathbf{j}}
\end{aligned}
\)

Hint

(d) From the question, \(P Q=2 r \sin \alpha=\frac{2 m v_0}{B q} \sin \alpha\) \(\because\)
\(
\theta=(2 \pi-2 \alpha)
\)
\(
\text { Here, } \alpha=\beta \text { and } v=v_0
\)

\(
\therefore \quad t=\frac{\theta}{\omega}=\frac{2(\pi-\alpha)}{(B q / m)}=\frac{2 m(\pi-\alpha)}{B q}
\)

Hint

(b)

Deflecting torque due to current,
\(
\tau_1=(n i A) B \sin 90^{\circ}=n i L^2 B
\)
Restoring torque,
\(
\begin{aligned}
\tau_2 & =\text { force } \times \text { perpendicular distance }=M g \times \frac{L}{2} \\
\tau_1 & \geq \tau_2 \\
n i L^2 B & \geq M g \frac{L}{2} \Rightarrow B \geq \frac{M g}{2 n i L} \\
B_{\min } & =\frac{M g}{2 n i L}
\end{aligned}
\)

Note \(\boldsymbol{M}\) is along \(+Z\)-axis and \(\boldsymbol{B}\) is along \(+X\)-axis. The magnetic moment vector \(\boldsymbol{M}\) tries to align along \(\boldsymbol{B}\) through smaller angle to have minimum potential energy, so coil rotates clockwise about line \(B C\).

Hint

\(
\text { (a) In the } z=0 \text { plane, the situation is as follows }
\)

Here, \(P(x, y)\) is the point and \(r=\sqrt{x^2+y^2}\) Magnetic field at \(P\) is perpendicular to \(O P\), as it is shown in figure.

\(
\begin{aligned}
& \text { So, } \quad \mathbf{B}=B \sin \theta \hat{\mathbf{i}}-B \cos \theta \hat{\mathbf{j}} \\
& \text { Here, } \quad B=\frac{\mu_0}{2 \pi} \frac{I}{r} \\
& \sin \theta=\frac{y}{r}, \cos \theta=\frac{x}{r} \\
& \therefore \quad \mathbf{B}=\frac{\mu_0 I}{2 \pi r}\left(\frac{y}{r} \hat{\mathbf{i}}-\frac{x}{r} \hat{\mathbf{j}}\right)=\frac{\mu_0}{2 \pi} \frac{I(y \hat{\mathbf{i}}-x \hat{\mathbf{j}})}{\left(x^2+y^2\right)}
\end{aligned}
\)

Hint

(c) Given, radius of identical circular loops,
\(
R=10 \mathrm{~cm}=10^{-1} \mathrm{~m}
\)
Current, \(I=\frac{7}{2} \mathrm{~A}\)
Distance from both circular loops at point \(P\),
\(
x_1=x_2=x=5 \mathrm{~cm}=5 \times 10^{-2} \mathrm{~m}
\)
From figure, according to Maxwell’s right hand thumb rule, it is clear that magnetic field will be in same direction by both the coils,
\(
\begin{aligned}
& \text { i.e. } \quad B=B_1+B_2=\frac{\mu_0 I R^2}{2\left(R^2+x^2\right)^{3 / 2}}+\frac{\mu_0 I R^2}{2\left(R^2+x^2\right)^{3 / 2}} \\
& =\frac{\mu_0 I R^2}{\left(R^2+x^2\right)^{3 / 2}}=\frac{\mu_0 \times 7 / 2 \times\left(10^{-1}\right)^2}{\left[(0.1)^2+(0.05)^2\right]^{3 / 2}} \\
& =\frac{56}{\sqrt{5}} \mu_0 \mathrm{~T}
\end{aligned}
\)

Hint

(a) Given, velocity of proton, \(\mathbf{v}=2 \hat{\mathbf{i}}\)
Magnetic field, \(\mathbf{B}=(\hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}) \mu \mathrm{T}\)
Electric field, \(\mathbf{E}=10 \hat{\mathbf{i}} \mu \mathrm{Vm}^{-1}\)
Applied Lorentz force on the proton,
\(
\begin{aligned}
\mathbf{F} & =q \mathbf{E}+q(\mathbf{v} \times \mathbf{B})=q[\mathbf{E}+(\mathbf{v} \times \mathbf{B})] \\
& =1.6 \times 10^{-19}\left[10 \times 10^{-6} \hat{\mathbf{i}}+2 \hat{\mathbf{i}} \times(\hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}) \times 10^{-6}\right] \\
& =1.6 \times 10^{-19} \times 10^{-6}[10 \hat{\mathbf{i}}+6 \hat{\mathbf{k}}-8 \hat{\mathbf{j}}] \\
\mathbf{F} & =1.6 \times 10^{-25}[10 \hat{\mathbf{i}}-8 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}] \mathrm{N}
\end{aligned}
\)
\(\therefore\) Acceleration of proton,
\(
\begin{aligned}
\mathbf{a} & =\frac{\mathbf{F}}{m_p} \quad\left(\because m_p=1.6 \times 10^{-27} \mathrm{~kg}\right) \\
& =\frac{1.6 \times 10^{-25}[10 \hat{\mathbf{i}}-8 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}]}{1.6 \times 10^{-27}} \\
\mathbf{a} & =100[10 \hat{\mathbf{i}}-8 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}] \\
\therefore \quad a & =|\mathbf{a}|=100 \sqrt{10^2+8^2+6^2} \\
& =100 \times 14.14=1414 \mathrm{~ms}^{-2} \simeq 1400 \mathrm{~ms}^{-2}
\end{aligned}
\)

Hint

(a) When an electron enters into a uniform magnetic field in a direction perpendicular to the direction of magnetic field, then a magnetic force acts on the electron in perpendicular direction of both direction of magnetic field and direction of velocity of electron and direction of force can be determined by Fleming’s left hand rule.
We know that, when direction of force on a particle is in perpendicular direction to the direction of velocity, then particle moves in uniform circular motion. Therefore, electron moving perpendicular to magnetic field (B) will perform circular motion.
Hence, both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

Hint

(a) Given, speed of proton, \(v=4.5 \times 10^5 \mathrm{~ms}^{-1}\)
Electrostatic force between two protons,
\(
F_e=k \frac{e^2}{r^2} \dots(i)
\)
Magnetic field produced due to moving proton of speed \(v\),
\(
B=\frac{\mu_0}{4 \pi} \cdot \frac{e v}{r^2} \dots(ii)
\)
\(\therefore[latex] Magnetic force on the proton,
[latex]
\begin{aligned}
& F_m=B e v=\frac{\mu_0}{4 \pi} \cdot \frac{e v}{r^2} \cdot e v \text { [From Eq. (ii)] }\\
& F_m=\frac{\mu_0}{4 \pi} \cdot \frac{e^2 v^2}{r^2} \dots(iii)
\end{aligned}
\)
From Eqs. (i) and (iii), we get
\(
\begin{aligned}
\frac{F_e}{F_m} & =\frac{\frac{k e^2}{r^2}}{\frac{\mu_0}{4 \pi} \cdot \frac{e^2 v^2}{r^2}}=\frac{k \cdot 4 \pi}{v^2 \cdot \mu_0} \\
\frac{F_e}{F_m} & =\frac{9 \times 10^9 \times 4 \times 3.14}{\left(4.5 \times 10^5\right)^2 \times 4 \pi \times 10^{-7}}=4.4 \times 10^5
\end{aligned}
\)

Hint

\(
\text { (a) } \because B(\text { at } P)=\frac{\mu_0 I}{4 \pi d}\left(\cos \theta_1-\cos \theta_2\right)
\)

\(
\begin{array}{r}
\text { In given case, } d=R \sin 45^{\circ}=R / \sqrt{2} \\
\theta_1=135^{\circ}, \theta_2=180^{\circ}
\end{array}
\)
\(
\begin{aligned}
\therefore B(\operatorname{at} P) & =\frac{\mu_0 I}{4 \pi(R / \sqrt{2})}\left[\cos 135^{\circ}-\cos 180^{\circ}\right] \\
& =\frac{\mu_0 I}{4 \pi R} \sqrt{2}\left[\frac{-1}{\sqrt{2}}-(-1)\right] \\
& =\frac{\mu_0 I}{4 \pi R} \sqrt{2}\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right) \\
\Rightarrow B(\operatorname{at} P) & =\frac{\mu_0 I}{4 \pi R}(\sqrt{2}-1) \mathrm{T}
\end{aligned}
\)

Hint

(b)
\(
\begin{aligned}
\begin{aligned}
d l & =d x=1 \mathrm{~cm}=10^{-2} \mathrm{~m}, \\
i & =10 \mathrm{~A}, r=0.5 \mathrm{~m} \\
\therefore \quad d B & =\frac{\mu_0}{4 \pi} \cdot \frac{i(d l \times r)}{r^3} \\
& =\frac{\mu_0}{4 \pi} \cdot \frac{i d l}{r^2}(\hat{\mathbf{i}} \times \hat{\mathbf{j}}) \\
& =\frac{\mu_0}{4 \pi} \cdot \frac{i d l}{r^2} \hat{\mathbf{k}} \\
& =\frac{10^{-7} \times 10 \times 10^{-2}}{(0.5)^2} \hat{\mathbf{k}}=4 \times 10^{-8} \hat{\mathbf{k}} \mathrm{~T}
\end{aligned}
\end{aligned}
\)

Hint

(a) Magnetic moment,
\(
M=N i A
\)

where, \(N=\) number of turns in the current loop
and \(\quad i=\) current.
Since, the orbiting electron behaves as a current loop of current \(i\), we can write
\(
i=\frac{e}{T}=\frac{e}{2 \pi r / v}=\frac{e v}{2 \pi r}
\)

\(
\begin{aligned}
&\text { where, } A=\text { area of the loop }=\pi r^2\\
&\Rightarrow \quad M=(1)\left(\frac{e v}{2 \pi r}\right)\left(\pi r^2\right)=\frac{e v r}{2}
\end{aligned}
\)

Hint

(a) The torque acting on a current carrying loop is given by
\(
\begin{aligned}
\tau & =M B \cos \theta \\
& =N i A B \cos \theta [\because M=N i A]
\end{aligned}
\)
where, \(N=\) number of turns,
\(i=\) current of loop,
\(A=\) area of loop
and \(B=\) magnetic field.
Thus, torque does not depend on shape of loop.

Hint

(a) According to Biot-Savart’s law, the magnetic induction at a point to a current carrying element \(i \delta l\) is given by
\(
B=\frac{\mu_0}{4 \pi} \cdot \frac{i \delta l \sin \theta}{r^2}
\)
Directed normal to plane containing \(\delta l\) and \(\mathrm{r}, \theta\) being angle between \(\delta l\) and \(r\).
Field due to semicircular arc
Now, angle between a current element \(\delta l\) of semicircular arc and the radius vector of the element to point \(c\) is \(\pi / 2\).
Therefore, the magnitude of magnetic induction \(B\) at \(O\) due to this element,
\(
\delta B=\frac{\mu_0}{4 \pi} \cdot \frac{i \delta l \sin \pi / 2}{r^2}=\frac{\mu_0 i \delta l}{4 \pi r^2}
\)
Hence, magnetic induction due to whole semicircular loop is
\(
\begin{aligned}
B & =\Sigma \delta B=\Sigma \frac{\mu_0}{4 \pi} \cdot \frac{i \delta l}{r^2} \\
& =\frac{\mu_0}{4 \pi} \frac{i}{r^2} \Sigma \delta l=\frac{\mu_0 i}{4 \pi r^2}(\pi r)=\frac{\mu_0 i}{4 r}
\end{aligned}
\)
The magnetic field due to \(a b\) and \(d e\) is zero, because \(\theta=0^{\circ}\) or \(180^{\circ}\), so net magnetic field is
\(
B=\frac{\mu_0 i}{4 r}
\)

Hint

(c) Magnetic field due to straight wire above \(O\) is zero, i.e. \(B_1=0\) (since, \(\theta=0^{\circ}\) )
The magnetic field due to semicircular part,
\(
B_2=\frac{1}{2} \times \frac{\mu_0 I}{2 r}=\frac{\mu_0 I}{4 r}
\)
The magnetic field due to lower straight portion,
\(
B_3=\frac{\mu_0 I}{4 \pi r}\left(\sin 0^{\circ}+\sin 90^{\circ}\right)=\frac{\mu_0 I}{4 \pi r} \quad \text { (upward) }
\)
\(
\begin{aligned}
&\text { Net magnetic field, }\\
&B=B_1+B_2+B_3=0+\frac{\mu_0 I}{4 r}+\frac{\mu_0 I}{4 \pi r}
\end{aligned}
\)
\(
=\frac{\mu_0 I}{4 r}+\frac{\mu_0 I}{4 \pi r} \text { (upwards) }\\
\)

Hint

(b) The angle subtended by circular part of the conductor is \(3 \pi / 2[latex] or [latex]270^{\circ}\).
Net magnetic field at point \(O, B_{\text {net }}=B_1+B_2\) where, \(B_1=\) magnetic field due to arc II and \(B_2=\) magnetic field due to arc \(I\).
\(
B_{\text {net }}=\frac{\mu_0 i}{4 \pi \times 3 R} \times \frac{\pi}{2}+\frac{\mu_0 i}{4 \pi R} \times 3 \frac{\pi}{2}=\frac{5 \mu_0 i}{12 R} \quad \text { (downward) }\\
\)

Hint

(c) Given, speed of proton, \(v=3 \times 10^6 \mathrm{~m} / \mathrm{s}\)
Magnetic field, \(B=2 \times 10^{-3} \mathrm{~T}\)
Now, force (magnetic) on proton,
\(
\begin{aligned}
F & =q v B \sin \theta \\
& =1.6 \times 10^{-19} \times 3 \times 10^6 \times 2 \times 10^{-3} \times \sin 90^{\circ} \\
& =3.2 \times 3 \times 10^{-16}=9.6 \times 10^{-16} \mathrm{~N}
\end{aligned}
\)
\(\therefore\) Acceleration of proton,
\(
\begin{aligned}
a & =\frac{F}{m}=\frac{9.6 \times 10^{-16} \mathrm{~N}}{1.67 \times 10^{-27} \mathrm{~kg}}=\frac{9.6}{1.67} \times 10^{11} \mathrm{~m} / \mathrm{s}^2 \\
& =5.74 \times 10^{11} \mathrm{~m} / \mathrm{s}^2 \approx 5.8 \times 10^{11} \mathrm{~m} / \mathrm{s}^2
\end{aligned}
\)

Hint

(c) Magnetic field at the centre of circular coil,
\(
B=\frac{\mu_0 I}{2 r} \dots(i)
\)
Now magnetic field at the centre of smaller circular coil,
\(
B^{\prime}=\frac{\mu_0 n i}{2 r / n} \Rightarrow B^{\prime}=n^2 \frac{\mu_0 i}{2 r}
\)
From Eq. (i), we get
\(
B^{\prime}=n^2 B \Rightarrow \frac{B^{\prime}}{B}=\frac{n^2}{1}
\)

Hint

(a) Two infinitely long and straight parallel wires carrying equal currents in same direction will attract each other.

Hint

(b) Points \(A\) and \(A^{\prime}\) are the inflection points, where the sign of curvature changes.
At these points (i.e. \(A\) and \(A^{\prime}\) ), field is constant, so \(\frac{d B}{d t}\) as well as \(\frac{d^2 B}{d t^2}\) are zero.

Hint

(b) Let the masses of two particles are \(m_1\) and \(m_2\).
As the charges are of the same magnitude and being accelerated through same potential, so these charges enter into the magnetic field with the same speeds (let \(v\) ). Now, radii of the circular paths followed by two changes is given by
\(
R_1=\frac{m_1 v}{q B} \text { and } R_2=\frac{m_2 v}{q B} \Rightarrow \frac{R_1}{R_2}=\frac{m_1}{m_2}
\)

Hint

(c) The magnetic field produced by the ring at its centre,
\(
B=\frac{\mu_0 I}{2 r} \dots(i)
\)
where, \(I=\) current through the ring
\(
=\frac{Q}{T}=Q /(2 \pi / \omega)=\frac{Q \omega}{2 \pi} \dots(ii)
\)
Here, \(Q=\) total change on the ring
and \(\quad T=\) time period of the ring.
According to the question, \(\lambda=\frac{Q}{2 \pi r} \Rightarrow Q=\lambda \times 2 \pi r \dots(iii)\)
From Eqs. (i), (ii) and (iii), we get
\(
B=\frac{\mu_0 \lambda \omega}{2}
\)

Hint

(c) Since, \(r=\frac{\sqrt{2 m q V}}{q B}=\sqrt{\frac{2 m V}{q B^2}} \Rightarrow \frac{r_1}{r_2}=\sqrt{\frac{m_1}{m_2}} \Rightarrow \frac{m_1}{m_2}=\frac{r_1^2}{r_2^2}\)
Hence, \(\frac{m_1}{m_2}=\frac{(2)^2}{(3)^2}=\frac{4}{9}\)

Hint

(a) Given, \(B=10^{-4} \mathrm{~Wb} / \mathrm{m}^2\),
\(
\frac{q}{m}=10^{11} \mathrm{C} / \mathrm{kg} \text { and } v=10^9 \mathrm{~m} / \mathrm{s}
\)
Radius of the circle described,
\(
\begin{aligned}
& r=\frac{m v}{q B} \Rightarrow r=\frac{v}{\left(\frac{q}{m}\right) B} \\
& r=\frac{10^9}{\left(10^{11}\right)\left(10^{-4}\right)} \Rightarrow r=100 \mathrm{~m}
\end{aligned}
\)

Hint

(d) As, electrons cannot be accelerated in a cyclotron. A large increase in their energy increases their velocity to a very large extent. This throws the electrons out of step with the oscillating field while neutron, being electrically neutral, cannot be accelerated in a cyclotron.
So, cyclotron is used to accelerate only positively charged particles.

Hint

(c) Two parallel beams of positron moving in the same direction set up two parallel currents flowing in the same direction. Hence, they attract each other.

Hint

(a) Given, \(I_1=3 \mathrm{~A}, I_2=4 \mathrm{~A}, R=2 \pi \mathrm{~cm}=2 \pi \times 10^{-2} \mathrm{~m}\)
Magnetic field at the centre of a coil, \(B=\frac{\mu_0 I}{2 R}\)
Now, \(B_1=\frac{\mu_0 I_1}{2 R}=\frac{\mu_0}{2} \times \frac{3}{2 \pi \times 10^{-2}}=\frac{\mu_0}{4 \pi} \times \frac{3}{10^{-2}}\)
\(=10^{-7} \times \frac{3}{10^{-2}}=3 \times 10^{-5} \mathrm{~T}\)
Similarly, \(B_2=\frac{\mu_0 I_2}{2 R}=\frac{\mu_0}{2} \times \frac{4}{2 \pi \times 10^{-2}}=\frac{\mu_0}{4 \pi} \times \frac{4}{10^{-2}}\)
\(=10^{-7} \times \frac{4}{10^2}=4 \times 10^{-5} \mathrm{~T}\)
Now, net magnetic field at centre of the coils, \(B=\sqrt{B_1^2+B_2^2}\)
\(
\begin{array}{ll}
\Rightarrow & B=\sqrt{\left(3 \times 10^{-5}\right)^2+\left(4 \times 10^{-5}\right)^2} \\
\Rightarrow & B=\sqrt{10^{-10}(9+16)} \\
\Rightarrow & B=5 \times 10^{-5} \mathrm{~T}=5 \times 10^{-5} \mathrm{~Wb} / \mathrm{m}^2
\end{array}
\)

Hint

(b) The magnetic field inside the solenoid is given by
\(
B=\mu_0 n I
\)
where, \(n=\) number of turns per unit length
and \(\quad I=\) current in coil
\(
\text { Now, } \quad n=\frac{N}{L}=\frac{400}{0.4 \times 10^{-2}}=10^5
\)
and \(\quad \mu_0=4 \pi \times 10^{-7} \mathrm{~T}-\mathrm{m} / \mathrm{A}\)
\(
\begin{aligned}
\therefore \quad B=4 & \times 3.14 \times 10^5 \times 5 \times 10^{-7} \\
& =62800 \times 10^{-5}=6.28 \times 10^{-1} \mathrm{~T}
\end{aligned}
\)

Hint

(b) The magnetic field inside a toroid is given by \(B=\frac{\mu_0 N I}{2 \pi r}\).
Magnetic field \(\mathbf{B}\) at any point in the open space inside the toroid is zero because the amperian loop encloses net current equal to zero.

Hint

(b) The magnetic field at axis of current carrying loop is given by
\(
B=\frac{\mu_0 i r^2}{2\left(r^2+d^2\right)^{3 / 2}} \dots(i)
\)
where, \(i=\) current in loop, \(r=\) radius of loop \(d=\) distance on axis and \(\quad B=[latex] magnetic field of current carrying loop. Let [latex]B_1\) be the magnetic field at centre ( \(d=0\) ).
Now, \(\quad B_1=\frac{\mu_0 i r^2}{2\left(r^2+0^2\right)^{3 / 2}}=\frac{\mu_0 i r^2}{2 r^3}=\frac{\mu_0 i}{2 r} \dots(ii)\)
\(
\begin{aligned}
&\text { On dividing Eqs. (i) and (ii), we get }\\
&\frac{B}{B_1}=\frac{\mu_0 i r^2 \times 2 r}{2\left(r^2+d^2\right)^{3 / 2} \times \mu_0 i}
\end{aligned}
\)
\(
\begin{aligned}
& \frac{B}{B_1}=\frac{\mu_0 i r^2 \times 2 r}{2\left(r^2+d^2\right)^{3 / 2} \times \mu_0 i} \\
& \frac{B}{B_1}=\frac{r^3}{\left(r^2+d^2\right)^{3 / 2}}=\frac{3^3}{\left(3^2+4^2\right)^{3 / 2}} \\
& \frac{B}{B_1}=\frac{27}{(25)^{3 / 2}} \Rightarrow \frac{54 \mu \mathrm{~T}}{B_1}=\frac{27}{5^3} \\
& B_1=\frac{54 \times 125}{27} \\
& B_1=250 \mu \mathrm{~T}
\end{aligned}
\)

Hint

(b) Given, kinetic energy \(=E\)
\(
\begin{aligned}
& \qquad \text { Mass }=m \\
& \text { Magnetic field }=B \\
& \text { and charge }=q \\
& \text { Magnetic force, } F=q v B \sin \theta \\
& \text { If } \theta=90^{\circ} \text {, then } F=q v B \dots(i)
\end{aligned}
\)
Now, centripetal force,
\(
F=\frac{m v^2}{r} \dots(ii)
\)
From Eqs. (i) and (ii), we get
\(
q v B=\frac{m v^2}{r} \Rightarrow r=\frac{m v}{q B}
\)
\(
\begin{array}{ll}
\therefore & r=\frac{m \sqrt{\frac{2 E}{m}}}{q B} \\
\Rightarrow & r=\frac{\sqrt{2 E m}}{q B}
\end{array} \quad\left(\because E=\frac{1}{2} m v^2, v=\sqrt{\frac{2 E}{m}}\right)
\)

Hint

(d) Let the velocity \(v\) of the particle entering the field \(\mathbf{B}\), instead of being perpendicular to \(\mathbf{B}\) makes an angle with it.

Then, \(v\) may be resolved into two components: \(v_{\|}=v \cos \theta\) parallel to \(\mathbf{B}\) and \(v_{\perp}=v \sin \theta\) perpendicular to \(\mathbf{B}\). The component \(v_{\|}\)gives a linear path and the component \(v_{\perp}\) gives a circular path to the particle. The resultant of these two is a helical path whose axis is parallel to the magnetic field.

Hint

(c) The number of revolutions per second by the charge \(q\) is given as

\(
N=\frac{1}{T}=\frac{v}{2 \pi R}
\)
Current through circular path, \(i=q(1 / T)=v q / 2 \pi R\)
Magnetic moment, \(\mu=i A=\frac{v q}{2 \pi R}\left(\pi R^2\right) \Rightarrow \mu=\frac{1}{2} v q R\)

Hint

(d) Consider a charged particle of mass \(m\) and charge \(q\) moving along a circular path in anti-clockwise direction as shown in figure.

The magnetic field \(B\) is assumed to be into plane of paper and of constant magnitude.
Clearly, centripetal force required for the circular motion of the charge will be provided by magnetic Lorentz force \(q(v \times B)\).
Hence,
\(
\begin{aligned}
\frac{m v^2}{r} & =q v B \sin 90^{\circ} \\
\frac{m v^2}{r} & =q v B
\end{aligned}
\)
\(
\text { Speed, } v=\frac{q B r}{m}
\)
\(
r=\frac{m v}{q B} \dots(i)
\)
Let \(T\) be the time period of the periodic motion, i.e. time taken to complete one revolution.
\(
\begin{gathered}
\Rightarrow \quad T=\frac{2 \pi}{\omega}=\frac{2 \pi}{v / r}=\frac{2 \pi r}{v}=\frac{2 \pi}{v}\left(\frac{m v}{q B}\right) \text { [from Eq. (i)] }\\
=\frac{2 \pi m}{q B}
\end{gathered}
\)

Hint

(b) Magnetic moment,
\(
\begin{aligned}
M & =\text { Current } \times \text { Area of enclosed by loop } \\
& =I \times A \dots(i)
\end{aligned}
\)
Magnetic induction at the centre of circular loop,
\(
B=\frac{\mu_0 I}{2 R} \Rightarrow I=\frac{2 B R}{\mu_0} \dots(ii)
\)
Here, \(A=\pi R^2 \dots(iii)\)
Substituting Eqs. (ii) and (iii) in Eq. (i), we get
\(
M=\frac{2 B R}{\mu_0} \times \pi R^2=\frac{2 \pi B R^3}{\mu_0}
\)

Hint

\(
\begin{aligned}
&\text { (c) In cyclotron, force on charge }=\text { centripetal force }\\
&\therefore \quad q v B=\frac{m v^2}{r} \Rightarrow r=\frac{m v}{B q} \Rightarrow r \propto v
\end{aligned}
\)

Hint

(a) When a magnetic field is applied on a stationary electron, then it remains stationary.
Because,\(\mathbf{F}=q(\mathbf{v} \times \mathbf{B})\)
If \(v=0 \Rightarrow F=0\)

Hint

\(
\begin{aligned}
\text { (d) Given, } r=0.05 \mathrm{~nm}=0.05 \times 10^{-9} \mathrm{~m}, & \left(\because 1 \mathrm{~nm}=10^{-9} \mathrm{~m}\right) \\
n=10^{16} \mathrm{rev} / \mathrm{s}, e=1.6 \times 10^{-19} \mathrm{C} &
\end{aligned}
\)
\(
\begin{aligned}
&\text { Magnetic moment, } M=A i\\
&\begin{aligned}
M & =\pi r^2 \times n e \quad \quad\left(\because A=\pi r^2\right) \\
& =3.14 \times\left(0.05 \times 10^{-9}\right)^2 \times 10^{16} \times 1.6 \times 10^{-19} \\
\text { or } \quad & =1.26 \times 10^{-23} \mathrm{~A}-\mathrm{m}^2
\end{aligned}
\end{aligned}
\)

Hint

(c) Given, radius of coil, \(r=10 \mathrm{~cm}=10 \times 10^{-2} \mathrm{~m}\)
Number of turns, \(N=100\) turns, current in coil, \(i=1 \mathrm{~A}\)
Magnetic moment, \(M=\) ?
Now, \(M=N i A\), where \(A=\pi r^2\)
\(
\begin{aligned}
M= & N i \pi r^2=100 \times 1 \times 3.142 \times\left(10 \times 10^{-2}\right)^2 \\
& \quad=100 \times 1 \times 3.142 \times 100 \times 10^{-4}=3.142 \mathrm{~A}-\mathrm{m}^2
\end{aligned}
\)

Hint

(c) The given situation is shown in the figure.

Since, given wires are very large, hence magnetic field due to current in the first wire,

(c)
\(
B_1=\left[\frac{\mu_0}{2 \pi d} I_1\right]
\)
The field is perpendicularly inward.
Now, force on the second wire due to first wire,
\(
\begin{aligned}
& \mathbf{F}_{21}=I_2\left(\mathbf{1} \times \mathbf{B}_1\right) \quad \text { (towards left) } \\
& F_{21}=I_2 l B_1 \sin 90^{\circ} \\
& F_{21}=I_2 l B_1
\end{aligned}
\)
where, \(l\) is the length of the wire and \(d\) is separation between the two wires.
\(
F_{21}=I_2 l\left(\frac{\mu_0 I_1}{2 \pi d}\right)=\frac{\mu_0 I_1 I_2 l}{2 \pi d}
\)
According to question, \(\left|\mathbf{F}_{21}\right|=F\)
\(
\Rightarrow \quad F=\frac{\mu_0 I_1 I_2 l}{2 \pi d}
\)
Now, consider the situation, when the current in first wire is doubled and reversed, we can write

\(
F^{\prime}=\frac{\mu_0\left(2 I_1\right)\left(I_2\right) l}{2 \pi d \times 3}
\)
Here, force on second wire will be towards right.
So,
\(
\begin{aligned}
& \frac{F^{\prime}}{F}=\left[\frac{\frac{\mu_0\left(2 I_1\right)\left(I_2\right) l}{2 \pi d \times 3}}{\frac{\mu_0 I_1 I_2 l}{2 \pi d}}\right] \\
& \frac{F^{\prime}}{F}=\left[\frac{2 I_1 I_2}{3 I_1 I_2}\right]=\frac{2}{3} \\
& F^{\prime}=-\frac{2 F}{3}
\end{aligned}
\)
\(
\Rightarrow \quad F^{\prime}=-\frac{2 F}{3}
\)
Here, negative sign is due to opposite directions of \(F^{\prime}\) and \(F\).

Hint

(a) From Fleming’s left-hand rule, when a particle is in motion and magnetic field is perpendicular to the velocity.

Hint

(d) Angular momentum, \(L=m v r \dots(i)\)
The orbital motion of electron is equivalent to a current.
\(
\therefore \quad I=e(1 / T)
\)
Period of revolution of electron, \(T=\frac{2 \pi r}{v}\)
\(
\therefore \quad I=e\left(\frac{1}{2 \pi r / v}\right)=\frac{e v}{2 \pi r}
\)
Area of electron orbit, \(A=\pi r^2\)
Magnetic dipole moment of the atom,
\(
M=I A=\frac{e v}{2 \pi r} \times \pi r^2=\frac{e v r}{2}
\)
Using Eq. (i), we have
\(
\begin{aligned}
M & =\left(\frac{e}{2 m}\right) L \\
\Rightarrow \quad \frac{M}{L} & =\frac{e}{2 m}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (c) Circumference of circle }=\text { Length of wire }\\
&\therefore \quad 2 \pi R=L
\end{aligned}
\)
\(
\Rightarrow \quad R=\left(\frac{L}{2 \pi}\right)
\)
Area, \(A=\pi R^2=\frac{\pi L^2}{4 \pi^2}=\frac{L^2}{4 \pi}\)
Magnetic moment, \(M=I A=\frac{I L^2}{4 \pi}\)

Hint

(d) The given figure can be drawn as follows.

\(
\begin{aligned}
&\text { The magnetic field at point } P \text {, }\\
&\begin{aligned}
B_{\mathrm{net}} & =2\left[\frac{\mu_0 I}{4 \pi r}\left(\sin \theta_1+\sin \theta_2\right)\right] \\
& =2\left[\frac{\mu_0}{4 \pi} \times \frac{I}{\frac{d \sqrt{3}}{2}} \times\left(\sin 90^{\circ}+\sin 30^{\circ}\right)\right] \\
& =2\left[\frac{\mu_0}{4 \pi} \times \frac{2 I}{d \sqrt{3}} \times\left(1+\frac{1}{2}\right)\right] \\
& =2\left[\frac{\mu_0}{4 \pi} \times \frac{2 I}{d \sqrt{3}} \times \frac{3}{2}\right]=\frac{\sqrt{3} \mu_0 I}{2 \pi d}
\end{aligned}
\end{aligned}
\)

Hint

(a) The radius of the circular path of a charged particle in magnetic field,
\(
r=\frac{m v}{q B}=\frac{\sqrt{2 m E}}{q B}
\)
Here, kinetic energy for proton and helium is same and both are moving in the same magnetic field.
\(
\begin{aligned}
r & \propto \frac{\sqrt{m}}{q} \\
\frac{r_P}{r_{\mathrm{He}}} & =\frac{\frac{\sqrt{m_P}}{q_P}}{\frac{\sqrt{m_{\mathrm{He}}}}{q_{\mathrm{He}}}}=\sqrt{\frac{m_P}{m_{\mathrm{He}}}} \times \frac{q_{\mathrm{He}}}{q_P} \\
& =\sqrt{\frac{m}{4 m}} \cdot \frac{2 q}{q}=\frac{1}{1}
\end{aligned}
\)

Hint

\(
\text { (a) Time period, } T=\frac{2 \pi m}{B q}
\)
\(
\begin{aligned}
T & =\frac{m}{q} \\
\frac{T_1}{T_2} & =\frac{m_1}{m_2} \times \frac{q_2}{q_1}=\frac{1}{4} \times \frac{3}{2}=\frac{3}{8} \text { or } 3: 8
\end{aligned}
\)

Hint

(c) The magnetic field at centre of a coil, \(B=\frac{\mu_0 N i}{2 R}\)
Given, \(\quad i=2 \mathrm{~A}, N=1\)
So, \(\quad R=\frac{\mu_0 N i}{2 B}\)
\(
\begin{aligned}
R & =\frac{4 \pi \times 10^{-7} \times 1 \times 2}{2 \times 4 \pi \times 10^{-6}} \\
R & =\frac{1}{10}=0.1 \mathrm{~m}
\end{aligned}
\)

Hint

(d) Given that, the net magnetic field at the centre \(O\) is zero. Therefore, magnetic field at \(O\) due to circular coil and straight conductor must be equal and opposite in direction.
\(
\therefore \quad \frac{\mu_0 I_1}{2 R}=\frac{\mu_0 I_2}{2 \pi(2 R)} \Rightarrow \frac{I_1}{I_2}=\frac{1}{2 \pi}
\)

Hint

(c) The charged particle goes undeflected through both the fields, therefore force experience by charged particle due magnetic field must be equal to the force experienced by the charge particle due to electric field, i.e. \(F_m=F_e\) or \(e v B \sin \theta=e E\)
Given, \(v=2 \times 10^3 \mathrm{~ms}^{-1}\)
\(
B=1.5 \mathrm{~T} \text { and } \theta=90^{\circ}
\)
Hence,
\(
\begin{aligned}
E & =v B \sin \theta=2 \times 10^3 \times 1.5 \times \sin 90^{\circ} \\
& =3 \times 10^3 \mathrm{~V} / \mathrm{m} \text { or } \mathrm{N} / \mathrm{C}
\end{aligned}
\)

Hint

(a) The magnetic field \((B)\) at the centre of circular current carrying coil of radius \(R\) and current
\(
I, B=\frac{\mu_0 I}{2 R}
\)
Similarly, if current \(=2 I\), then
\(
\text { Magnetic field }=\frac{\mu_0 2 I}{2 R}=2 B
\)
So, resultant magnetic field
\(
\begin{aligned}
& =\sqrt{B^2+(2 B)^2} \\
& =\sqrt{5 B^2}=\sqrt{5} B \\
& =\frac{\mu_0 I \sqrt{5}}{2 R}
\end{aligned}
\)

Hint

\(
\text { (d) The given figure is shown below }
\)
Magnetic field outside the long rod,
\(
B=\frac{\mu_0 i}{2 \pi r} \quad\left[\text { for rod } 3 \text { and rod } 4, r_3<r_4\right]
\)
\(
\therefore \quad B_3>B_4 \neq 0
\)

Hint

(d) The magnetic field at the centre of new loop
\(
B^{\prime}=n^2 B
\)
where, \(n\) is the number of turns in the loop.
\(
\therefore \quad B^{\prime}=(4)^2 B \text { or } B^{\prime}=16 B
\)

Hint

(c) Under uniform magnetic field, force \(e v B\) acts on proton and provides the necessary centripetal force \(\frac{m v^2}{a}\).
\(
\therefore \quad \frac{m v^2}{a}=e v B \Rightarrow v=\frac{a e B}{m}
\)
Now, angular momentum,
\(
L=r \times p=a \times a e B=a^2 e B
\)

Hint

(c) Given, \(B=8.35 \times 10^{-2} \hat{\mathbf{i}} \mathrm{~T}, v=\left(2 \times 10^5 \hat{\mathbf{i}}+4 \times 10^5 \hat{\mathbf{j}}\right) \mathrm{m} / \mathrm{s}\)
The distance covered by proton, \(d=T(v)=2 \pi \frac{m}{q B}(v)\)
\(
\begin{aligned}
= & 2 \times 3.14 \times \frac{1.67 \times 10^{-27}}{1.6 \times 10^{-19} \times 8.35 \times 10^{-2} \hat{\mathbf{i}}} \\
& \times\left(2 \times 10^5 \hat{\mathbf{i}}+4 \times 10^5 \hat{\mathbf{j}}\right) \\
\Rightarrow \quad d= & 0.157 \mathrm{~m}
\end{aligned}
\)

Hint

(d) Given, \(n=12\) turns, \(I=15 \mathrm{~A}, B=0.2 \hat{\mathbf{i}}\),
\(
A=-0.04 \hat{\mathbf{i}} \mathrm{~m}^2
\)
\(\therefore\) Potential energy, \(U=n I A B\)
\(
=12 \times 15 \times(-0.04) \times 0.2=-1.44 \mathrm{~J}
\)

Hint

(b) The described condition can be shown as

The magnetic field at \(P\) due to inner and outer conductors are equal and opposite. Hence, the net magnetic field at \(P\) will be zero.

Hint

(b) When the currents in the wires are in same direction. Magnetic field at mid-point \(O\) due to I and II wires are respectively
\(
B_1=\frac{\mu_0}{4 \pi} \frac{2 i_1}{x} \otimes \text { and } B_{\mathrm{II}}=\frac{\mu_0}{4 \pi} \cdot \frac{2 i_2}{x} \odot
\)

So, the net magnetic field at \(O\),
\(
\begin{aligned}
B_{\text {net }} & =\frac{\mu_0}{4 \pi} \cdot \frac{2}{x}\left(i_1-i_2\right) \\
\Rightarrow \quad 10 \times 10^{-6} & =\frac{\mu_0}{4 \pi} \cdot \frac{2}{x}\left(i_1-i_2\right) \dots(i)
\end{aligned}
\)
When the direction of \(i_2\) is reversed,
\(
B_{\mathrm{I}}=\frac{\mu_0}{4 \pi} \cdot \frac{2 i_1}{x} \otimes
\)
and
\(
B_{\mathrm{II}}=\frac{\mu_0}{4 \pi} \cdot \frac{2 i_2}{x} \otimes
\)

\(
\begin{aligned}
&\text { So, net magnetic field at } O \text {, }\\
&\begin{aligned}
B_{\text {net }} & =\frac{\mu_0}{4 \pi} \cdot \frac{2}{x}\left(i_1+i_2\right) \\
\Rightarrow \quad 40 \times 10^{-6} & =\frac{\mu_0}{4 \pi} \cdot \frac{2}{x}\left(i_1+i_2\right) \dots(ii)
\end{aligned}
\end{aligned}
\)
\(
\begin{aligned}
&\text { On dividing Eq. (ii) by Eq. (i), we get }\\
&\frac{i_1+i_2}{i_1-i_2}=\frac{4}{1} \Rightarrow \frac{i_1}{i_2}=\frac{5}{3}
\end{aligned}
\)

Hint

(a) Because magnetic force always points perpendicular to the particle velocity. That is why velocity remains unchanged thereby keeping energy \(\frac{1}{2} m v^2\) and momentum (\(mv\)) unchanged.

Hint

(c) When wire is bent in the form of semicircular arc, then

\(
l=\pi r
\)
\(\therefore\) The radius of semicircular arc, \(r=l / \pi\)
Distance between two end points of semicircular wire
\(
=2 r=\frac{2 l}{\pi}
\)
\(\therefore\) Magnetic moment of semicircular wire
\(
=m \times 2 r=m \times \frac{2 l}{\pi}=\frac{2}{\pi} m l
\)
But \(m l\) is the magnetic moment of straight wire, i.e. \(m l=M\)
\(\therefore\) New magnetic moment \(=\frac{2}{\pi} M\)

Hint

\(
\begin{aligned}
&\text { (c) Radius, }\\
&\begin{aligned}
r & =\frac{m v}{q B} \\
B & =\frac{m v}{q r}=\frac{9.1 \times 10^{-31} \times 1.3 \times 10^6}{1.6 \times 10^{-19} \times 0.35} \\
& =2.1 \times 10^{-5} \mathrm{~T}
\end{aligned}
\end{aligned}
\)

Hint

(a) The magnetic field induction at \(P\) due to currents through both the wires,

\(
B=\frac{\mu_0}{4 \pi} \frac{2 i}{r / 2}+\frac{\mu_0}{4 \pi} \frac{2(2 i)}{r / 2}=\frac{\mu_0}{4 \pi} \cdot \frac{12 i}{r}
\)
It is acting perpendicular to plane of wire inwards. Now, \(B\) and \(v\) are acting in the same direction, i.e. \(\theta=0^{\circ}\).
\(\therefore\) Force on charged particle is \(F=q v B \sin \theta=q v B \times 0=0\).

Hint

(b) We have, \(M=N I A, B=\mu_0 n I\)
Torque, \(\tau=M B\)
Here,
\(
\begin{aligned}
\tau & =\left(n_1 I_1 A\right)\left(\mu_0 n_2 I_2\right) \\
& =\left(10 \times \frac{21}{44} \times 10^{-6}\right)\left(4 \times \frac{22}{7} \times 10^{-7} \times 10^3 \times 2.5\right) \\
& =1.5 \times 10^{-8} \mathrm{~N}-\mathrm{m}
\end{aligned}
\)