Magnetic Effects Of Current And Magnetism – JEE Main PhysicsPractice Questions & PYQs

Given, the charge on the proton, $q_{p}=+e$ The charge on the alpha particles, $q_{a}=+2 {e}$ The expression of the radius of the orbit, $r=\;\frac{m v}{q B}$ $\Rightarrow r=\;\frac{p}{q B}$ $( \because p=m v)$ $\because \;\; \;\frac{r_{p}}{r_{\alpha }}=\;\frac{p_{p}}{q_{p}} \times \;\frac{q_{\alpha }}{p_{\alpha }}$ $\Rightarrow \;\; \;\frac{2}{1}=\;\frac{+2 e}{+e} \times \;\frac{P_{p}}{p_{\alpha }}$ $[ \because \;\frac{r_{\rho }}{r_{\alpha }}=\;\frac{2}{1}]$ $\Rightarrow \;\; \;\frac{p_{p}}{p_{a}}=1$ Since, kinetic energy and momentum are related as $K=\;\frac{p^{2}}{2 m}$ $\therefore \;\frac{K_{p}}{K_{\alpha }}=\;\frac{p_{p}^{2} m_{\alpha }}{m_{p} p_{\alpha }^{2}} \Rightarrow \;\frac{K_{p}}{K_{\alpha }}=\;\frac{m_{\alpha }}{m_{p}}=\;\frac{4 m_{p}}{m_{p}}$ $\Rightarrow \;{K_{ {p}}}/{K_{\alpha }}=\;\frac{4}{1}$

Given, number of turns = N Inner radius = a Outer radius = b Current = I Consider an element of thickness dr at a distance r from the centre.Now, number of turns in dr element, $\text{dN}=\frac{\text{N}}{(b-a)} \text{dr}$ Magnetic field due to this element at the centre, $\text{dB}=\frac{\\mu _{0}}{2} \frac{\text{dN} .\text{I}}{r}$ $\therefore \text{dB}=\frac{1}{2} \frac{\\mu _{0} \text{I}}{r} \frac{\text{N}}{(b-a)} \text{dr}$ ∴ Total magnetic field at the centre, $\text{B}=\int _{a}^{b} \text{dB}=\int _{a}^{b} \frac{\\mu _{0} \text{IN}}{2(b-a)} \frac{dr}{r}$ $=\frac{\text{N}}{2} \frac{\\mu _{0} \text{I}}{(b-a)} \ln r|_{a} ^{b}$ $=\frac{\text{N}}{2} \frac{\\mu _{0} \text{l}}{(b-a)} \ln (\frac{b}{a})$

The magnetic field intensity at a distance r from centre on the axis of a coil of radius a carrying current i is given by $\text{B}_{{\text{axis} }}=\frac{\\mu _{0}}{4 \pi } \frac{2 \pi i a^{2}}{(a^{2}+r^{2})^{3 / 2}}$ ...(i) The magnetic field intensity at the centre of the coil is given by $\text{B}_{{\text{centre} }}=\frac{\\mu _{0}}{4 \pi } \frac{2 \pi i}{a}$ ...(ii) According to question, the fractional change in the magnetic fieldintensity is given by $\frac{\Delta\text{B}}{\text{B}}=1-{\text{B}_{{\text{axis} }}}/{\text{B}_{{\text{centre} }}}$ ...(iii) Substituting the values of $\text{B}_{{\text{axis} }}$ and $\text{B}_{{\text{centre} }}$ from Eqs. (i) and (ii) respectively in Eq. (iii), we get $\frac{\Delta\text{B}}{\text{B}}=1-{\frac{\\mu _{0}}{4 \pi } \frac{2 \pi i a^{2}}{(a^{2}+r^{2})^{3 / 2}}}/{\frac{\\mu _{0}}{4 \pi } \frac{2 \pi i}{a}}$ $=1-\frac{a^{3}}{(a^{2}+r^{2})^{3 / 2}}$ $=1-{a^{3}}/{a^{3}(1+\frac{r^{2}}{a^{2}})^{3 / 2}}$ $=1-(1+\frac{r^{2}}{a^{2}})^{-3 / 2}$ Given that, r < a, applying Binomial expansion, $(1-x)^{-n}=1-x+\frac{2 x^{2}}{2 !}-\frac{3 \times 2 x^{3}}{3 !} ...(-1)^{n} x^{n},$ $=(1+n x)$ we get $\frac{\Delta\text{B}}{\text{B}}=1-(1-\frac{3}{2} \frac{r^{2}}{a^{2}})$ $\frac{\Delta\text{B}}{\text{B}}=1-1+\frac{3}{2} \frac{r^{2}}{a^{2}}$ $\frac{\Delta\text{B}}{\text{B}}=\frac{3}{2} \frac{r^{2}}{a^{2}}$, which is the required expression.

Given, number of turns, $n=1000 {\text{m}}^{-1}$ Electric current, i = 0.75 A Relative permeability of 1st material $\mu_{1}=500$ Relative permeability of 2nd material $\mu_{2}=750$ we know that $\mu_{r}=1+\chi \Rightarrow \chi$ $=\mu_{r}-1$ For 1st and 2nd material, $\chi _{1}=\chi _{1}-1=500-1=499$ and $\chi _{2}=\mu_{2}-1=750-1=749$ Now, magnetic intensity, M = ni Let $\text{M}_{1}$ and $\text{M}_{2}$ be the magnetisation and $m_{1}$ and $m_{2}$ be the magnetic moments of material 1st and 2nd. Using, $\frac{\text{I}_{1}}{\text{M}}=\chi _{1}$ ...(i) and $\frac{\text{I}_{2}}{\text{M}}=\chi _{2}$ ...(ii) Dividing Eqs. (i) and (ii), $\frac{\text{I}_{1}}{\text{I}_{2}}=\frac{\chi _{1}}{\chi _{2}}$ ...(iii) We also know that, $\text{Magnetisation}={{ \text{Dipole} \text{moment} \in \text{core} \text{material} }}/{ { \text{Volume} }}$ i.e $\text{I}=\frac{m}{\text{V}}$ $\frac{\text{I}_{1}}{\text{I}_{2}}=\frac{m_{1}}{m_{2}}$ ...(iv) From Eqs. (iii) and (iv), we get $\frac{m_{1}}{m_{2}}=\frac{\chi _{1}}{\chi _{2}}=\frac{499}{749}$ Hence, fraction change in magnetic moment, $\frac{\Deltam}{m_{1}}=\frac{m_{2}-m_{1}}{m_{1}}=(\frac{m_{2}}{m_{1}}-1)$ $=(\frac{749}{499}-1)=\frac{250}{499}$ Thus, value of χ is 250.

According to question, a charge $Q$ is moving $d$ distance in magnetic field $B$. As we know that force acting on a point charge by magnetic field is always perpendicular to the velocity of particle $v$. $\therefore F=Q(v \times B)$.....(i) The above, statement means that ${F}$ is perpendicular to $v$. $\because$ Work done, $W=F \cdot d s$ $=Q(v \times B) \cdot d l$ [using Eq. (i)] $=Q(v \times B) d l \cos \theta$ $=Q(v \times B) d l \cos 90^{\circ}$ $[ \because \theta =90^{\circ} {F} ⟂ v$ or ${dl}]$ $=0$ $\therefore$ The work done by the magnetic field will be zero.

In the given figure, The direction of magnetic moment is from South to North. When second magnet is kept in the magnetic field produced by firstmagnet, the torque acting on second magnet is given by $\tau ={\text{M}}_{2} \times {\text{B}}_{1}$ $|\tau |=\text{M}_{2} \text{B}_{1} \sin \theta$ (∴ θ = 90°) $|\tau |=\text{M}_{2} \text{B}_{1}$ Point P lie on equitorial position, so $\text{B}_{1}=\frac{\\mu _{0}}{4 \pi } \frac{\text{M}_{1}}{r^{3}}$ $|\tau |=(\frac{\\mu _{0}}{4 \pi } \frac{\text{M}_{1}}{r^{3}}) \text{M}_{2}$ $\text{M}_{1}=\text{M}_{2}=1 {\text{A}}-{\text{m}}^{2}, r=1 {\text{m}}$ (given) $|\tau ||=\frac{10^{-7} \times 1 \times 1}{(1)^{3}}$ $\Rightarrow |\tau |=10^{-7} {\text{Nm}}$ $\Rightarrow |\tau |=1 \times 10^{-7} {\text{Nm}}$

A loop of wire will act like magnetic dipole. In equilibrium, direction of dipole will be parallel to direction of magnetic field and its plane remains normal to the field.Now on each element of wire force due to magnetic field will be perpendicular to length of element and outward in direction. So, loop will become circular. Hence, the correct option is (a).

Given, magnetic field at the centre at $A$ is $3 T$. Since, the solenoid $B, C$ and $D$ are connected in parallel arrangement. We know that, The magnetic field in the circuit is directly proportional to the $B \propto i$ $\Rightarrow \;\; \;\frac{B_{A}}{B_{C}}=\;\frac{I_{A}}{I_{C}} \Rightarrow \;\frac{3 T}{B_{C}}=\;\frac{i}{i / 3}$ $\Rightarrow \;\; B_{C}=1 {T}$ So, the field at the centre at $C=1 T$.

Given, mass of 1 st ion, $m_{1}=4 {\text{amu}}$ Mass of 2 nd ion, $m_{2}=16 {\text{amu}}$ Charge on 1 st ion, $q_{1}=+2 {e}$ Charge on 2 nd ion, $q_{2}=+3 {e}$ Kinetic energy of both ion is same, ${\text{KE}}_{1}={\text{KE}}_{2}$ The radius of path traced by a charged particle in magnetic field is given as $r={\\sqrt{2 \text{mKE}}}/{\text{qB}}$ $\Rightarrow r \propto\frac{\\sqrt{m}}{q}$ $\Rightarrow \frac{r_{1}}{r_{2}}=\\sqrt{\frac{m_{1}}{m_{2}}} \times (\frac{q_{2}}{q_{1}})$ $\Rightarrow \frac{r_{1}}{r_{2}}=\\sqrt{\frac{4}{16}} \times (\frac{+3 e}{2 e})$ $=\frac{1}{2} \times \frac{3}{2}=\frac{3}{4}$ $\Rightarrow r_{1}=\frac{3}{4} r_{2}$ $\Rightarrow r_{1} < r_{2}$ Now, the deflection of ion in region with constant magnetic field as shown in figure is denoted by θ. From diagram, $\sin \theta =\frac{d}{r}$ $\theta \propto\frac{1}{r}$ [\because d = constant] As, $r_{1} < r_{2}$ from above expression $\theta '_{1} > \theta _{2}$. Thus, deflection of lighter ion will be more than the deflection of heavier ion.

Side of equilateral triangular coil, a = 10 cm = 0.1 m Horizontal magnetic field, $\text{B}=20 {\text{mT}}=20 \times 10^{-3} {\text{T}}$ Current in coil, i = 0.2 A Torque acting on coil, $\tau =\\sqrt{x} \times 10^{-5} {\text{N}}-{\text{m}}$ Torque acting on coil can be calculated as τ = MBsin θ Here, M is pole strength and B is magnetic field.Magnetic pole strength, M = iA $\tau =i(\frac{\\sqrt{3}}{4} a^{2}) \text{B} \sin \theta$ $[$\because\text{A}=\frac{\\sqrt{3}}{4} a^{2}.$ (for equilateral triangle)]$ $=0.2 \times \frac{\\sqrt{3}}{4} \times (0.1)^{2} \times 20 \times 10^{-3} \times \sin 90°$ [\because A ⊥ B] $=\\sqrt{3} \times 10^{-5} {\text{N}}-{\text{m}}$ After comparing obtained value with the given value, x = 3.

Given, radius of inner wire of co-axial cable $=a$Innner radius of outer shell $=b$Outer radius of outer shell $=c$Let, magnetic field at distance $x< a=B_{1}$ and magnetic field at distance$a< x< b=B_{2}$By using Ampere circuital law,$∮ B \cdot d l=n \mu _{0} l_{\;\text{enc }\;}$where, $d l=$ perimeter of small circular element,$n=$ number of turns,$\mu _{0}=$ free space permeabilityand ${I}=$ current.$\therefore \;\; B_{1} \cdot 2 \pi x=n \mu _{0} I_{\;\text{enc }\;}$ $\Rightarrow \;\; B_{1} \;\; =\;\frac{\mu _{0}}{2 \pi x} \;\frac{I_{0}}{\pi a^{2}} \cdot \pi x^{2}$ $\;\; =\;\frac{\mu _{0}}{2 \pi } \;\frac{I_{0} x}{a^{2}}$Now, for $B_{2}(a< x< b)$ $B_{2} \cdot 2 \pi x \;\; =\mu _{0} I_{0}$ $B_{2} \;\; =\;\frac{1}{2 \pi } \;\frac{\mu _{0} I_{0}}{x}$ $\therefore$ From Eqs. (i) and (ii) we will get$\;\frac{B_{1}}{B_{2}}=\;{\;\frac{\mu _{0} I_{0} x}{2 \pi a^{2}}}/{\frac{\mu _{0} I_{0}}{2 \pi x}}=\;\frac{x^{2}}{a^{2}}$

Given, $d_{1}=0.05 {m}, d_{2}=0.2 {m}$ and $B_{1}: B_{2}=8: 1$ Let radius of coil be $r$. As we know that, magnetic field at a point on axis of a coil, $B=\;\frac{\mu _{0}}{2} \cdot \;\frac{I r^{2}}{(d_{1}^{2}+r^{2})^{3 / 2}}$ $\; \therefore \;\;\frac{B_{1}}{B_{2}} \;\; =\;\frac{r^{2} \cdot (d_{2}^{2}+r^{2})^{3 / 2}}{(d_{1}^{2}+r^{2})^{3 / 2} \cdot r^{2}}=\;\frac{(d_{2}^{2}+r^{2})^{3 / 2}}{(d_{1}^{2}+r^{2})^{3 / 2}}$ $\;\;\frac{8}{1}=(\;\frac{d_{2}^{2}+r^{2}}{d_{1}^{2}+r^{2}})^{3 / 2} \Rightarrow \;\frac{2^{3}}{1}=(\;\frac{d_{2}^{2}+r^{2}}{d_{1}^{2}+r^{2}})^{3 / 2}$ $\;\;\frac{4}{1}=\;\frac{d_{2}^{2}+r^{2}}{d_{1}^{2}+r^{2}}$ $\Rightarrow \;4 d_{1}^{2}+4 r^{2} \;\; =d_{2}^{2}+r^{2}$ $\Rightarrow \;4(0.05)^{2}+4 r^{2} \;\; =(0.2)^{2}+r^{2} \Rightarrow 3 r^{2}=0.03$ $\Rightarrow \;r=0.1 {m}$

The magnetic field $B$ produced by a circular arc at its center is given by the formula $B = \frac{\mu_0 I \theta}{4 \pi R}$, where $I$ is the current, $R$ is the radius, and $\theta$ is the subtended angle. For two wires $A$ and $B$, the ratio of their magnetic fields is expressed as:

$\frac{B_A}{B_B} = \frac{I_A}{R_A} \cdot \frac{R_B}{I_B} \cdot \frac{\theta_A}{\theta_B}$

Substituting the given values $I_A = 2 \text{ A}$, $R_A = 2 \text{ cm}$, $I_B = 3 \text{ A}$, and $R_B = 4 \text{ cm}$:

$\frac{B_A}{B_B} = \left( \frac{2}{2} \right) \cdot \left( \frac{4}{3} \right) \cdot \frac{\theta_A}{\theta_B} = \frac{4}{3} \cdot \frac{\theta_A}{\theta_B}$

To satisfy the target ratio of $6:5$, we equate $\frac{4}{3} \cdot \frac{\theta_A}{\theta_B} = \frac{6}{5}$, which implies the ratio of the subtended angles is $\frac{\theta_A}{\theta_B} = \frac{18}{20} = \frac{9}{10}$. Based on these parameters, the calculated ratio of the magnetic fields is $6:5$.

$\vec{\tau }={\vec{M}} \times {\vec{B}}$ $=4 {a}^{2} {I} \times \frac{\mu_{0} {I}}{2 \pi {b}}$

${F}={BI} 2 {a}=\frac{\mu_{0} {I}}{2 \pi {r}} {I} \times 2 {a}$ ${F}={\mu_{0} {I}^{2} {a}}/{\pi \sqrt{{b}^{2}+{a}^{2}}}$ $\tau ={F} \cos \theta \times 2 {a}$ $={\mu_{0} {I}^{2} {a}}/{\pi \sqrt{{b}^{2}+{a}^{2}}} \times {{b}}/{\sqrt{{b}^{2}+{a}^{2}}} \times 2 {a}$ $\tau =\frac{2 \mu_{0} {I}^{2} {a}^{2} {b}}{\pi ({a}^{2}+{b}^{2})}$ If ${b}>> a$ then $\tau =\frac{2 \mu_{0} {I}^{2} {a}^{2}}{\pi {b}}$ But among the given options (1) is most appropriate.

Top view of solenoidMaximum possible radius of electron $= R/2$ $\therefore R/2=\frac{mv}{qB}=\frac{mv_{\max}}{e(\mu _0ni)}$ $v_{\max}=R/2\frac{e\mu _0ni}{m}$

Pitch $= \frac{2 \pi {m}}{\ {qB}} \ {v} \cos \theta$ Pitch $= \frac{2(3.14) (1.67 \times 10^{-27} ) \\times 4 \\times 10^{5} \\times \cos 60}{\ (1.69 \\times 10^{-19} )(0.3)}$ Pitch $=0.04 {m}=4 {cm}$

In uniform magnetic field particle moves in a circular path, if the radius of the circular path is 'd', particle will not hit the screen. ${d}={{\text{m}\upsilon }}/{{\text{qB}}_{0}}$ ${\upsilon }={{\text{qB}}_{0} {\text{d}}}/{{m}}$

$a={qvB}/m$ $B=\frac{ma}{qv}={ma\text{sqrtm}}/\sqrt{2k}$ $=\frac{m^{3/2}a}{e\sqrt{2k}}$ $={(1.6\times 10^{-27})^{3/2}\times 10^12}/{1.6\times 10^{-19}\sqrt{2\times 1\times 10^6\times 1.6\times 10^{-19}}}$= 0.71 mT

$\vec{T}=\vec{M}\times \vec{B}\sin\theta =-MB\sin\theta$ $I\alpha =-MB\sin\theta$For small θ, $\alpha =-{MB}/I\theta$ $\omega =\sqrt{{MB}/I}=\sqrt{ {(i)(\pi R^2)B} /{ ({MR^2}/2 )}}$ $\omega =\sqrt{ {2i\pi B}/m }$ $\therefore T={2\pi }/\omega =\sqrt{ \frac{2\pi m}{iB} }$