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

Magnetic field inside cylinder at distance $r$ from axis$\;=\;\frac{\mu _0 I}{\pi R^2} \cdot \;\frac{r}{2}$ $B_{\max }\;=\;\frac{\mu _0 I}{2 \pi R}$ $\;\frac{\mu _0 I}{2 \pi R^2} r_1\;=\;\frac{1}{2} (\;\frac{\mu _0 I}{2 \pi R})$ $r_1\;=\;\frac{R}{2}=\;\frac{a}{2}$Magnetic field outside cylinder at distance $r$ from axis$=\;\;\frac{\mu _0 I}{2 \pi r}$ $\;\frac{\mu _0 I}{2 \pi r_2}\;=\;\frac{1}{2} (\;\frac{\mu _0 I}{2 \pi R})$ $r_2=2 R\;=2 a$

$R_{A B C}=r\;R_{A D C}=2 r$ $i_1=\;\frac{2 l}{3}\;i_2=\;\frac{l}{3}$ $B_{\;\text{centre }\;}=\;{2 (\mu _0 {\sqrt{2}}) }/{4 \pi (\;\frac{a}{2}) } [\;\frac{2 l}{3}-\;\frac{l}{3}] ={\sqrt{2}} \;\frac{\mu _0 l}{3 \pi a}$

$\;K \theta =N I A B$ $\;\;\frac{\theta }{I}=\;\frac{N A B}{K}$ $\;\;\frac{\theta }{V}=\;\frac{\theta }{R I}=\;\frac{N A B}{K R}$ $\;\;\text{ Ratio of voltage sensitivity }\;= (\;\frac{N_1 A_1 B_1}{R_1}) (\;\frac{R_2}{N_2 A_2 B_2})$ $\;=\;\frac{5}{7} \times \;\frac{2}{1} \times \;\frac{1}{2} \times \;\frac{7}{5}=1$

$\;B=\;\frac{\mu _0 I \times 2}{4 \pi ℓ} \times 2 \cos 45^{\circ} \times 4$ $\;\Rightarrow B=\;{4 \pi \times 10^{-7} \times {\sqrt{2}}}/{4 \pi \times 1} \times 4 \times 4 \times \;{1}/{{\sqrt{2}}} \times 5$ $\;\Rightarrow B=8 \times 10^{-6} T$

$B=\;\frac{\mu _0 I}{2 r} \;\frac{\theta }{2 \pi } \;\;$ for an arcFor semicircle $B=\;\frac{\mu _0 l}{4 r}$ $B_{ {\text{net}}}\;=\;\frac{\mu _0 l}{4\{4 \pi \}}-\;\frac{\mu _0 l}{4\{6 \pi \}}$ $\;=\;\frac{4_0 l}{4 \pi } \{\;\frac{1}{4}-\;\frac{1}{6}\}$ $\;=10^{-7} \times 12 \times \;\frac{2}{24}=10^{-7} {\text{T}} \Rightarrow k=1$

$\;B v q=E q$ $\;E=B v$ $\;r=\;\frac{v m}{B q}$ $\;B=\;\frac{m v}{r q}$ $\;E= (\;\frac{m v}{r q}) v=\;\frac{m v^2}{r q}$ $\;=\;\frac{1.6 \times 10^{-27} \times 4 \times 10^{10}}{2 \times 10^{-2} \times 1.6 \times 10^{-19}}$ $\;=2 \times 10^4 {\text{N}} / {\text{C}}$ $\;x=2$

Let the velocity of the particle make an angle $\theta$ with initial direction when it is at a distance $x$.$\Rightarrow \;\frac{d x}{d t}=-v_0 sin \;\theta$ $\;\text{ Also, }\; \;\frac{d \theta }{d t}=\;\frac{q}{m} (\;\frac{\mu _0 I}{2 \pi x})$ $\Rightarrow \;\frac{d x}{d \theta }=\;\frac{-2 \pi m v_0 x sin \;\theta }{q \mu _0 I} . . .$ dividing eq(i) with eq(ii)$\;\text{ or }\; ∫_{a}^{x} \;\frac{d x}{x}=\;\frac{-2 \pi m v_0}{q \mu _0 I} ∫_{0}^{\pi } sin \;\theta d \theta$ $\ln \;\frac{x}{a}=\;\frac{-4 \pi m v_0}{q \mu _0 I}$Or $\;\; x=a e^{\;\frac{-4 \pi m v_0}{q \mu _0 l}}$

The refractive index $n$ of the medium is defined as the ratio of the speed of light in a vacuum ($c$) to its speed in the medium ($v$), which is given by the relation $n = \frac{c}{v} = \sqrt{\mu_r \epsilon_r}$.
Substituting the given relative permeability $\mu_r = \frac{10}{\pi}$ and relative permittivity $\epsilon_r = \frac{1}{0.0885}$, we get:
$n = \sqrt{\left( \frac{10}{\pi} \right) \left( \frac{1}{0.0885} \right)}$
Using the approximation $\pi \approx 3.1416$:
$n = \sqrt{\frac{10}{3.1416 \times 0.0885}} \approx \sqrt{\frac{10}{0.278}} \approx \sqrt{35.97}$
Evaluating the square root, we find $n \approx 5.997 \approx 6$.
Thus, the velocity of light is $6$ times greater in vacuum than in the medium.

$I=\;\frac{N q \omega }{2 \pi }$Now for bigger Amperian loop (loop B)$I_{B(\;\text{ enclosed }\;)}=0$And for smaller Amperian loop (loop $A$ )${| I_{A(\;\text{ enclosed }\;)} |}=I=\;\frac{N q \omega }{2 \pi }$So, ${| I_A-I_B |}=\;\frac{N q \omega }{2 \pi }$

$R=\;\frac{m v}{q B}$(A) $m v, B$ are same for {\text{O}}^{2-} & {\text{H}}^{+}therefore $R \propto \;\frac{1}{|q|}$ therefore (A) is correct(B) $R=\;\frac{m v}{q B} ; m v, q, B$ are same both therefore (R) is incorrect

$\;\text{ We know, }\;$ $\;\beta =\mu _0 η I \;\text{ and }\; T=\;\frac{2 \pi m}{e \beta }$ $\;\Rightarrow \;\frac{2 \pi m}{e T}=\mu _0 η I$ $\;\Rightarrow \;\; η=\;\frac{2 \pi m}{e T \mu _0 I}$ $\; \;\; =\;\frac{2 \pi \times 9 \times 10^{-31}}{1.6 \times 10^{-19} \times 75 \times 10^{-9} \times 4 \pi \times 10^{-7} \times 1.5}$ $\;\Rightarrow \;\; η=\;\frac{9 \times 10^{-31}}{1.6 \times 10^{-19} \times 75 \times 10^{-9} \times 2 \times 10^{-7} \times 1.5}$ $\;\Rightarrow \;\; η=\;\frac{9 \times 10^4}{360}=250 \;\text{ per meter }\;$

If the electron's velocity is along the direction of magnetic field then magnetic force on electron is zero and it will not accelerate.

We know inside the wire$B=\;\frac{\mu _0 I}{2 \pi R^2} \cdot r \;\; (O< r< R)$And $B=\;\frac{\mu _0 I}{2 \pi r} \;\;$ for $(R< r)$

$\;m v r=\;\frac{n h}{2 \pi } ......(i)$ $\;r=\;\frac{v m}{B q} ......(ii)$ $\;n=2$ $\;m r (\;\frac{r B q}{m}) =\;\frac{2 h}{2 \pi }$ $\;r=\sqrt{\;\frac{h}{\pi B q}}$ $\;q=e$ $\;r=\sqrt{\;\frac{h}{\pi B e}}$

$\;B_1=\;\frac{\mu _0 i}{2 R}$ $\;x=\;\frac{3 R}{4}$ $\;B_2=\;\frac{\mu _0 i R^2}{2 (R^{2+}x^2) ^{3 / 2}}=\;\frac{\mu _0 i R^2}{2 (\;\frac{5 R}{4}) ^3}$ $\;=\;\frac{64}{125} (\;\frac{\mu _0 i}{2 R})$ $\;\;\frac{B_2}{B_1}=\;\frac{64}{125}$

$\;\frac{1}{2} m v^2=E \Rightarrow (m v)^2=2 m E$Also, $r=\;\frac{m v}{q B} \Rightarrow r=\;{{\sqrt{2 m E}}}/{q B}$So, $r \propto (E)^{\;\frac{1}{2}}$ Graph should be like

$E\;= (\;\frac{B^2}{2 \mu }) \times \;\text{ Volume }\;$ $\mu \;=2 \mu _0$ $\;=\;\frac{B^2 A l}{4 \mu _0}$

At $P$ $\;\vec{B}=\vec{B}_1+\vec{B}_2$ $\;=\;\frac{\mu _0 i_2}{2 \pi r_2} \hat{k}-\;\frac{\mu _0 i_1}{2 \pi r_1} \hat{k}$ $\;=\;\frac{\mu _0}{2 \pi } (\;\frac{4}{4} \times 10^{2-}\;\frac{5}{10} \times 10^2) \hat{k}$ $\;=\;\frac{2 \times 10^{-7} \times 10^2}{2}=10^{-5}$ $\;x=1$

$\;B_{\;\text{net }\;}=\;\frac{\mu _0 I}{4 \pi a}+\;\frac{3}{8} \;\frac{\mu _0 I}{a}$ $\;=\;\frac{\mu _0 I}{4 \pi a} (1+\;\frac{3 \pi }{2})$

$\;R_1=\;\frac{m v}{q B_1}$ $\;R_2=\;\frac{m v}{q B_2}$ $\;|\Delta r|={| 2 R_2-2 R_1 |}=\;\frac{2 m v}{q} [\;\frac{1}{B_2}-\;\frac{1}{B_1}]$ $\;\Rightarrow \Delta r={| \;\frac{2 m v}{q B_1} [\;\frac{B_1}{B_2}-1] |}={| \;\frac{2 m v}{q B_1} (1-\;\frac{B_1}{B_2}) |}$