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

$\; {\text{B}}=4 \times \;{\mu _0 {\text{i}}}/{4 \pi (1 / 2)} (\;{1}/{{\sqrt{2}}}+\;{1}/{{\sqrt{2}}})$ $\;=4 \times 10^{-7} \times 5 \times 2 \times {\sqrt{2}}$ $\;=40 {\sqrt{2}} \times 10^{-7} {\text{T}}$

$\;B= (\;\frac{\mu _0 i}{2 \pi a}) \times 2$ $=\;\frac{4 \pi \times 10^{-7} \times 10}{\pi \times (\;\frac{5}{2} \times 10^{-2}) }$ $\;=16 \times 10^{-5}=160 \mu {\text{T}}$

$\; {\text{B}}_{ {\text{A}}}=\;{\mu _0 {\text{i}}}/{2 \pi {\text{r}}}+\;{\mu _0(2 {\text{i}})}/{2 \pi (3 {\text{r}})}=\;{5 \mu _0 {\text{i}}}/{6 \pi {\text{r}}}$ $\; {\text{B}}_C=\;{\mu _0(2 {\text{i}})}/{2 \pi {\text{r}}}+\;{\mu _0 {\text{i}}}/{2 \pi (3 {\text{r}})}=\;{7 \mu _0 {\text{i}}}/{6 \pi {\text{r}}}$ $\; \therefore \;{ {\text{B}}_A}/{ {\text{B}}_C}=\;\frac{5}{7}$ $\; \therefore {\text{x}}=5$

We know${\text{W}}_{\;\text{ext }\;}\;=\Delta {\text{U}}+\Delta {\text{KE}} \;\; (\;\text{ P.E. }\;=-{ {\text{M}}}^{\to } \cdot { {\text{B}}}^{\to })$ $\;=-{ {\text{M}}}^{\to } \cdot { {\text{B}}}^{\to }_{ {\text{f}}}+{ {\text{M}}}^{\to } \cdot { {\text{B}}}^{\to }_{ {\text{i}}}+0$ $\;=- {\text{MB}} \cos 90+ {\text{MB}} \cos 0$ $\;= {\text{MB}}$ $\;= {\text{NIAB}}$ $\;=200 \times 100 \times 10^{-6} \times \;\frac{5}{2} \times 10^{-4} \times 1=5 \mu {\text{J}}$

$\; \because {\text{V}}_0- {\text{V}}_{ {\text{A}}}=\;{ {\text{B}} \omega ℓ^2}/{2}$ $\; {\text{V}}_0- {\text{V}}_{ {\text{B}}}=\;{ {\text{B}} \omega ℓ^2}/{2}$ $\; \therefore {\text{V}}_{ {\text{A}}}= {\text{V}}_{ {\text{B}}} \therefore {\text{V}}_{ {\text{A}}}- {\text{V}}_{ {\text{B}}}=0$

Force on segment parallel to ${\text{x}}$-axis will cancel each other. Hence ${\text{F}}_{\;\text{net }\;}$ will be due to portion parallel to y-axis.$\;F=0.5 \times 0.5 \times 6 \times 0.2-0.5 \times 0.5 \times 0.2 \times 5$ $\;=0.5 \times 0.5 \times 0.2$ $\;=0.25 \times 0.2$ $\;=50 \times 10^{-3} {\text{N}}$ $\;=50 {\text{mN}}$

Net force on particle must be zero i.e. ${\text{q}} { {\text{E}}}^{\to }+ {\text{q}} { {\text{V}}}^{\to } \times { {\text{B}}}^{\to }=0$ Possible cases are (i) { {\text{E}}}^{\to } \ &\ { {\text{B}}}^{\to }=0 (ii) ${ {\text{V}}}^{\to } \times { {\text{B}}}^{\to }=0, { {\text{E}}}^{\to }=0$ (iii) ${\text{q}} { {\text{E}}}^{\to }=- {\text{q}} { {\text{V}}}^{\to } \times { {\text{B}}}^{\to }$ { {\text{E}}}^{\to } \ne 0\ &\ { {\text{B}}}^{\to } \ne 0

$\; {\text{B}}=6 (\;{\mu _0 {\text{I}}}/{4 \pi {\text{r}}}) ( sin \;30^{\circ}+ sin \;30^{\circ})$ $\;=6 \;{10^{-7} \times 4 \pi {\sqrt{3}}}/{ (\;{{\sqrt{3}} \times 4 \pi }/{2 \times 6}) }$ $\;=72 \times 10^{-7} {\text{T}}$

$\because B_{n e t}=\;{{\sqrt{2}} \mu _0 I}/{2 a}$

$\; {\text{W}}=\Delta {\text{U}}= {\text{U}}_{ {\text{f}}}- {\text{U}}_{ {\text{i}}}$ $\; {\text{W}}=(-\vec{\mu } \cdot { {\text{B}}}^{\to })_{ {\text{f}}}-(-\vec{\mu } . { {\text{B}}}^{\to })_{ {\text{i}}}$ $\;=0+(\vec{\mu } \cdot { {\text{B}}}^{\to })_{ {\text{i}}}$ $\;= (100 \times 5 \times 10^{-3} \times 1 \times 10^{-3}) \times 0.2 {\text{J}}$ $\;=1 \times 10^{-4} {\text{J}}=100 \mu {\text{J}}$

$\;{ {\text{F}}}^{\to }= {\text{q}}({ {\text{v}}}^{\to } \times { {\text{B}}}^{\to })$ \;5 {\text{e}} { {\text{k}}^{\^}}= {\text{e}}(3 { {\text{i}}^{\^}}+5 { {\text{j}}^{\^}}) \times ( {\text{B}}_0 { {\text{i}}^{\^}}+2 {\text{B}}_0 { {\text{j}}^{\^}}) \;5 {\text{e}} { {\text{k}}^{\^}}= {\text{e}} (6 {\text{B}}_0 { {\text{k}}^{\^}}-5 {\text{B}}_0 { {\text{k}}^{\^}}) $\;\Rightarrow {\text{B}}_0=5 {\text{T}}$

Note : Direction of magnetic field is in +{ {\text{k}}^{\^}} $\;{ {\text{F}}}^{\to }= {\text{i}} \vec{ℓ} \times { {\text{B}}}^{\to }$ $\;ℓ=2 {\text{R}}$ \;{ {\text{F}}}^{\to }=-2 {\text{iRB}} { {\text{j}}^{\^}}

$\; {\text{F}}_{ {\text{m}}}= {\text{mg}}$ $\; \therefore \;\; \;\text{ ILB }\;= {\text{mg}}$ $\; \therefore \;\; (\;{ {\text{V}}}/{ {\text{R}}}) {\text{LB}}= {\text{mg}}$ $\; \therefore \;\; {\text{V}}=\;{ {\text{mgR}}}/{ {\text{LB}}}$ $\;=\;{ (1 \times 10^{-3} {\text{kg}}) (10 {\text{m}} / {\ s}^2) (10 Ω)}/{(0.1 {\text{m}}) (10^3 \times 10^{-4} {\text{T}}) }=10 {\text{V}}$

Force per unit length between two parallel straight$\;\;\text{ wires }\;=\;{\mu _0 {\text{i}}_1 {\text{i}}_2}/{2 \pi {\text{d}}}$ $\;\;\frac{F_1}{F_2}=\;{\;{\mu _0(10)^2}/{2 \pi (5 {\text{cm}})}}/{{\mu _0(20)^2}/{2 \pi (\;{5 {\text{cm}}}/{2}) }}=\;\frac{1}{8}$ $\;\Rightarrow {\text{F}}_2=8 {\text{F}}_1$ $\;$

${\text{M}}=$ NIA$\; {\text{M}}_{ {\text{A}}}= {\text{M}}_{ {\text{B}}}$ $\therefore \; {\text{N}}_{ {\text{A}}} {\text{I}}_{ {\text{A}}} {\text{A}}_{ {\text{A}}}= {\text{N}}_{ {\text{B}}} {\text{I}}_{ {\text{B}}} {\text{A}}_{ {\text{B}}}$ $\therefore \; {\text{N}}_{ {\text{A}}} {\text{I}}_{ {\text{A}}} \pi (0.1)^2= {\text{N}}_{ {\text{B}}} {\text{I}}_{ {\text{B}}} \pi (0.2)^2$ $\therefore \; {\text{N}}_{ {\text{A}}} {\text{I}}_{ {\text{A}}}=4 {\text{N}}_{ {\text{B}}} {\text{I}}_{ {\text{B}}}$

\;{B_{\;\text{net }\;}}^{\to }=\;\frac{\mu _0 i}{2 r} {𝑖}^{\^}+\;\frac{\mu _0 i}{2 r} {𝑗}^{\^} $\;\Rightarrow B_{\;\text{net }\;}=\;\frac{\mu _0 i}{2 r} {\sqrt{2}}=4 \pi \times 10^{-7} \times {\sqrt{2}} \times {\sqrt{2}} \times \;\frac{1}{2 \times 0.2}=2 \times 3.14 \times 10^{-6}=628 \times 10^{-8} {\text{T}}$

$\tau = {\text{K}} \theta$ ${\text{NiAB}}= {\text{K}} \theta$ ${\text{A}}=\;{ {\text{K}} \theta }/{ {\text{NiB}}}=\;\frac{4 \times 10^{-5} \times 0.05}{200 \times 10 \times 10^{-3} \times 0.01}$On solving $A=10^{-4} {\text{m}}^2=1 {\text{cm}}^2$

$\; {\text{d}} \tan 60^{\circ}=2 {\sqrt{3}}$ $\; {\text{d}}=2 {\text{cm}}$ $\; {\text{B}}=3 \times \;{\mu _0 {\text{i}}}/{2 \pi {\text{d}}} sin \;60^{\circ}$ $\;=3 \times \;\frac{2 \times 10^{-7} \times 2}{2 \times 10^{-2}} \times \;{{\sqrt{3}}}/{2}$ $\;=3 {\sqrt{3}} \times 10^{-5}$

It is a case of solid infinite current carrying wire. Using ampere circuital law,CASE I: if ${\text{r}} \le {\text{R}}$ $B=\;\frac{\mu _0 i}{2 \pi R^2} r$CASE II: $r > {\text{R}}$ $B=\;\frac{\mu _0 i}{2 \pi r}$

$\; {\text{B}}_{\;\text{total }\;}= {\text{B}}_{ {\text{I}}}+ {\text{B}}_{ {\text{II}}}+ {\text{B}}_{\;\text{III }\;}$ $\; {\text{B}}_{ {\text{I}}}=0$ $\; {\text{B}}_{\;\text{III }\;}=0$Because $\vec{d l} \times \vec{r}=0$Now, magnetic field due to semicirclaur ring at its center is given by$\; {\text{B}}_{\;\text{II }\;}=\;{\mu _0 {\text{i}}}/{4 {\text{R}}}$ $\;=\;\frac{4 \pi \times 10^{-7} \times 14}{4 \times 2.2 \times 10^{-2}}$ $\;=\;\frac{22}{7} \times \;\frac{10^{-7} \times 14}{22 \times 10^{-3}}$ $\;=2 \times 10^{-7}$ $\;=2$