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

$m g \times \;\frac{1}{\sqrt{2}}=\;\frac{i l B}{\sqrt{2}}$ $\Rightarrow i=\;\frac{m g}{B l}$ $=\;\frac{0.45 \times 10}{0.15}=30 {\text{A}}$

Inside a hollow cylindrical conductor with uniform current distribution net magnetic field is zero in hollow space.But outside the cylindrical conductor $B \propto \;\frac{1}{r}$ $\Rightarrow$ Graph in option D would be a correct one

$\;\text{ Field at }\; {\text{P}} \;\text{ is }\;= (\;\frac{\mu _0 \times i_1}{2 \pi r_1}-\;\frac{\mu _0 i_2}{2 \pi r_2}) (-\hat{k})$ $=- (\;\frac{\mu _0 4}{2 \pi \times 0.04}-\;\frac{\mu _0 \times 2}{2 \pi \times 0.06}) \hat{k}=-\;\frac{\mu _0 \times 200}{6 \pi } \hat{k}$ So, force $\vec{F}=\vec{q v} \times \vec{B}$ $=3 \pi (2 \hat{i}+3 \hat{j}) \times (- (\;\frac{\mu _0 \times 200}{6 \pi } \hat{k}) )$ $=3 \pi (\;\frac{200 \mu _0}{3 \pi } \hat{j}-\;\frac{100 \mu _0}{\pi } \hat{i})$ $=200 \mu _0 \hat{j}-300 \mu _0 \hat{i}$ $=4 \pi \times 10^{-5}(2 \hat{j}-3 \hat{i})$ So, $x=3$

$\;\frac{314}{100}=2 \pi {\text{R}} \;\; {\text{R}}=0.5 {\text{m}}$Magnetic Moment $=$ IA$=14 \times \pi {\text{R}}^2$ $=14 \times (3.14) \times \;\frac{1}{4}$ $=10.99 \approx 11.00$

In the width of $r_2-r_1$ total ${\text{n}}$ turns presents.$\therefore$ In 1 unit width $\;\frac{n}{r_2-r_1}$ turns presents.$\therefore$ In the width of $d$ r number of turns,$n^{'}=\;\frac{n}{r_2-r_1} \times d r$Magnetic field $( {\text{dB}})$ due to element of dr length is$d B=\;\frac{\mu _0 \times I \times n^{'}}{2 r}$ $=\;\frac{\mu _0 I}{2 r} \times \;\frac{n}{ (r_2-r_1) }$ $\therefore$ Total magnetic field due to entire coil is,$∫ d B=∫_{r_1}^{r_2} \;\frac{\mu _0 I}{2 r} \times \;\frac{n}{ (r_2-r_1) } d r$ $\Rightarrow B=\;\frac{\mu _0 I n}{2 (r_2-r_1) } ∫_{r_1}^{r_2} \;\frac{d r}{r}$ =\;\frac{\mu _0 I n}{2 (r_2-r_1) } \times \[\log _e^r]$ _{r_2}^{r_1}$$=;\frac{\mu _0 I n}{2 (r_2-r_1) } \times (\log _e^{r_2}-\log _e^{r_1})$$=;\frac{\mu _0 I n}{2 (r_2-r_1) } \times \log _e ;\frac{r_2}{r_1}$

$\;\frac{d F}{d l}=\;\frac{\mu _0 i_1 i_2}{2 \pi d}$So $\;\frac{2 \times 10^{-7} \times 5 \times 5}{d}=\;\frac{10^{-5}}{10 \times 10^{-2}}$ $d=\;\frac{2 \times 10^{-7} \times 5 \times 5}{10^{-4}}$ $=50 {\text{mm}}$ $=5 {\text{cm}}$

$\vec{F}=\vec{i l} \times \vec{B}$ $=i l B \ sin 60^{\circ}$ $=10 \times \;\frac{5}{100} \times 0.5 \times \;\frac{\sqrt{3}}{2}$ $=0.2165 {\text{N}}$

Magnetic force $\vec{F} ⟂ \vec{v}$ $\Rightarrow W_b=0$ $\Rightarrow \Delta K E=0$ and speed remains constant.

As $B_{\;\text{net }\;} \ne 0$ that is the wires are carrying current in opposite direction.$\;\frac{\mu _0 I \times 2}{2 \pi (4 \times 10^{-2}) }=30 \times 10^{-6} {\text{T}}$ $\Rightarrow I=\;\frac{30 \times 10^{-6}}{10^{-6}} {\text{A}}=30 {\text{A}} \;\text{ \in opposite direction. }\;$

Kinetic energy of electron in cyclotron$=[\;{ {\text{q}}^2 {\text{B}}^2 {\text{r}}_0^2}/{2 {\text{m}}}]$ $=18 {\text{MeV}}$

$B=\;\frac{\mu _0 N I}{2 R}$ $\;\frac{B_X}{B_Y}=\;\frac{N_x R_y}{N_y R_x}$ $=\;\frac{200 \times 20}{400 \times 20}=\;\frac{1}{2}$

For a wire with radius $R$ and current $I$, the magnetic field $B$ inside ($r < R$) is $B(r) = \frac{\mu_0 I r}{2 \pi R^2}$ (linear) and outside ($r > R$) is $B(r) = \frac{\mu_0 I}{2 \pi r}$ (inversely proportional).

The peak magnitude at the surface is $B_{max} = \frac{\mu_0 I}{2 \pi R}$. Given $a < b$, the wire with radius $a$ has a higher peak magnetic field than the wire with radius $b$ ($B_{max,a} > B_{max,b}$), and this peak occurs at an earlier radial position ($r=a < r=b$).

Comparing these requirements to the options, graph C correctly shows the curve for wire 'a' peaking higher and earlier than the curve for wire 'b', followed by a $1/r$ decay for both.

${R}=\frac{{mv}}{{qB}} \Rightarrow\frac{{R}_{1}}{{R}_{2}}={\frac{{mv}_{1}}{{q}_{1} {B}}}/{\frac{{mv}_{2}}{{q}_{2} {B}}}$ $=\frac{{v}_{1}}{{q}_{1}} \\times \frac{{q}_{2}}{{v}_{2}}=\frac{{q}_{2}}{{q}_{1}} \\times \frac{{v}_{1}}{{v}_{2}}$ $=\frac{2}{1} \\times (\frac{2}{3})=\frac{4}{3}$

Given, number of turns per unit length, $n=1000$ Relative magnetic permeability, $\mu _{t}=500$ Current flowing through solenoid, $I=5 {A}$ Magnetic permeability of free space, $\mu _{0}=4 \pi \times 10^{-7} \frac{H}{m}$ We know that, magnetic field inside the solenoid can be given as $B=\mu n I$ ....(i) where, $\mu =$ permeability of medium. As $\;\; \mu =\mu _{r} \mu _{0}$...(ii) From Eqs. (i) and (ii), we get $\therefore \;\; B=\mu _{r} \mu _{0} n l$ $=500 \times 4 \pi \times 10^{-7} \times 1000 \times 5$ $=100 \pi \times 10^{-7} \times 10^{5}$ $=\pi T$

Given, electric current, I = 1.5 A Side of triangle, $l=9 {\text{cm}}=9 \times 10^{-2} {\text{m}}$ Let OD = OE = OF = r $\text{AD}=\frac{l}{2}$ ${\text{In}} \Delta\text{AOD}, \frac{\text{OD}}{\text{AD}}=\tan 30°$ $\text{OD}=\text{AD} \tan 30°$ $r=\text{OD}=\frac{l}{2} \times \frac{1}{\\sqrt{3}}=\frac{l}{2 \\sqrt{3}}$ Magnetic field due to AB at O, $\text{B}_{1}=\frac{\mu{0}^\text{I}}{4 \pi r}[\sin 60°+\sin 60°]=\frac{\mu{0}^\text{I}}{4 \pi r} \times \frac{2 \\sqrt{3}}{2}$ $\Rightarrow \text{B}_{1}=\frac{\mu_{0} d \\sqrt{3}}{4 \pi r}$ (perpendicular inward) Now, total magnetic field at O due to all sides, $\text{B}=3 \text{B}_{1}=\frac{\mu_{0}}{4 \pi r} 3 \\sqrt{3}$ By substituting the values, we get $\text{B}=\frac{10^{-7} \times 1.5 \times 3 \\sqrt{3} \times 2 \\sqrt{3}}{9 \times 10^{-2}}=3 \times 10^{-5} {\text{T}}$ (inside the plane of triangle)

${r}=\frac{{mv}}{{qB}}={\\sqrt{2 {mk}}}/{{qB}}$ ${{r}_{{d}}}/{{r}_{\alpha }}=\\sqrt{{{m}_{{d}}}/{{m}_{\alpha }}} {{q}_{\alpha }}/{{q}_{{d}}}$ $=\\sqrt{\frac{2}{4}}(\frac{2}{1})=\\sqrt{2}$ Hence option (2).

Let $F_{p}, F_{d}$ and $F_{\alpha }$ be the forces and $v_{p}, v_{d}$ and $v_{\alpha }$ be the velocities of proton, deuteron and $\alpha -$particle, respectively. Since, $F=B q v$ On dividing and multiplying $F$ by $m$, we get $F=B q v \;\frac{m}{m}$ $\Rightarrow \;\; F=B \;\frac{q}{m} p \;\; ( \because p=m v)$ $\Rightarrow \;\; F \propto q / m \;\; ( \because p$ and $B$ are same) $\therefore \;\; F_{p}: F_{d}: F_{\alpha } \;\; =\;\frac{+q}{m}: \;\frac{+q}{2 m}: \;\frac{+2 q}{4 m}\;$ $\; \;\; =1: 1 / 2: 1 / 2=2: 1: 1$ $\;\;\text{ and }\;\; \;\; =B q v$ $\Rightarrow \;\; \propto F / q$ $\therefore \;\; v_{p}: v_{d}: v_{\alpha } \;\; =\;\frac{F_{p}}{q_{p}}: \;\frac{F_{d}}{q_{d}}: \;\frac{F_{\alpha }}{q_{\alpha }}=\;\frac{2 x}{q}: \;\frac{x}{q}: \;\frac{x}{2 q}$ $\;\; =2: 1: 1 / 2=4: 2: 1$

Length of conducting wire, l = 24 a Resistance of wire is R. Ratio of magnetic moment of coil for equilateral triangle to square is $1: \text{sqrty}$. Side of equilateral triangle is a and also side of square is a. Number of turns in equilateral triangle will be $\text{n}_{\text{T}}=\frac{24 a}{3 a}=8$ [ \because Length of one turn is 3a] Number of turns in square will be $\text{n}_{\text{S}}=\frac{24 a}{4 a}=6$ [ \because Length of one turn is 4a] Magnetic moment of coil is given as M = nIA The ratio of magnetic moment for triangular coil to square coil is given as ${\text{M}_{\text{T}}}/{\text{M}_{\text{S}}}={\text{n}_{\text{T}} \text{IA}_{\text{T}}}/{\text{n}_{\text{S}} \text{IA}_{\text{S}}}$ (\because I is same in both coils) Substituting the values in above expression, we get $\frac{1}{\\sqrt{y}}={(8 \times \frac{\\sqrt{3}}{4} a^{2})}/{6 \times a^{2}}=\frac{8 \\sqrt{3}}{24}=\frac{1}{\\sqrt{3}}$ $\Rightarrow \\sqrt{y}=\\sqrt{3}$ ⇒ y = 3 Hence, the value of y is 3.

The diagram is given below, The magnetic field due to the wire 1, $\text{B}_{{\text{wire} } 1}=\frac{\\mu _{0} \text{I}}{4 \pi y}(\sin 90°+\sin \theta _{1})$ $\frac{\\mu _{0} \text{I}}{4 \pi y} (1+x/\sqrt{x^{2+}y^2})$ The magnetic field due to the wire 2, $\frac{\\mu _{0} \text{I}}{4 \pi y}(\sin 90°+\sin \theta _{2})$ $\text{B}_{{\text{wire} } 2}=\frac{\\mu _{0} \text{I}}{4\pi x}(1+y/\sqrt{x^{2+}y^2})$ Thus, the total magnetic field, $\text{B}=\text{B}_{1}+\text{B}_{2}$ $=\frac{\\mu _{\0} \text{I}}{4 \pi }(\frac{1}{y}+{x}/{y \\sqrt{x^{2}+y^{2}}}+\frac{1}{x}+{y}/{x \\sqrt{x^{2}+y^{2}}})$ $=\frac{\\mu _{\0} \text{I}}{4 \pi xy}[\text{sqrtx}^{2+}y^{2+}(x+y)]$

For the straight wire, the expression of the magnetic field, $B_{1}=B_{3}=\;\frac{\mu _{0} I}{4 \pi r}$ For the semicircular wire, the expression of the magnetic field, $B_{2}=\;\frac{\mu _{0} I}{4 r}$ Hence, the total magnetic field at the point $P$, $B=B_{1}+B_{2}+B_{3}$ Substituting the values in the above equation, we get $B=\;\frac{\mu _{0} I}{4 \pi r}+\;\frac{\mu _{0} I}{4 r}+\;\frac{\mu _{0} I}{4 \pi r} \Rightarrow B=\;\frac{\mu _{0} I}{4 \pi r}(1+1+\pi )$ $B=\;\frac{\mu _{0} I}{4 \pi r}(2+\pi )$