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

The magnetic field at the center of a circular loop of radius $r$ carrying current $I$ is $B_{loop} = \frac{\mu_0 I}{2r}$. The magnetic field at a distance $r$ from an infinitely long straight wire is $B_{wire} = \frac{\mu_0 I}{2 \pi r}$. According to the principle of superposition, the net magnetic field at the center is the vector sum of these contributions. Given the configuration results in field subtraction, the magnitude is:
$B = \frac{\mu_0 I}{2r} - \frac{\mu_0 I}{2 \pi r} = \frac{\mu_0 I}{2 \pi r} \left( \pi - 1 \right)$
The negative sign in the provided option indicates the direction is opposite to the defined unit vector $\hat{i}$.

$\tau = | { \vec{\text{M}}} \times { \vec{\text{B}}} |$ $= | {\text{NI}} { \vec{\text{A}}} \times { \vec{\text{B}}} |$ $\tau =125 \times 1 \times \pi (\;\frac{2}{100}) ^{2} \times 0.4 \times \sin 30^{\circ}$ $=100 \pi \times 10^{-4} {\text{N}}- {\text{m}}$ $So\, \alpha =100 \pi =314$

$5 (\;{2 \pi {\text{m}}}/{ {\text{qB}}}) {\text{v}} \Rightarrow \;\frac{5 \times 2 \times \pi \times 5 \times 10^{-6} \times 2 \times 10^{-2}}{5 \pi \times 10^{-6} \times 0.1}=2 {\text{m}}$

$B_{1}=\;\frac{\mu _{0} I}{4 \pi r} \times 2+\;\frac{\mu _{0} I}{4 r}=\;\frac{\mu _{0} I}{4 r} (\;\frac{2}{\pi }+1)$ ${\text{B}}_{2}=\;{\mu _{0} {\text{I}}}/{4 \pi {\text{r}}}+\;{\mu _{0} {\text{I}}}/{4 {\text{r}}}=\;{\mu _{0} {\text{I}}}/{4 {\text{r}}} (\;\frac{1}{\pi }+1)$ $\;{ {\text{B}}_{1}}/{ {\text{B}}_{2}}=\;\frac{2+\pi }{1+\pi }$

Magnetic field at centre $B= (\;\frac{\mu _{0}}{4 \pi }) \;\frac{2 \pi i}{R}$ $=\;\frac{10^{-7} \times 2 \times \pi \times 2}{10 \times 10^{-2}}$So energy stored${\text{U}}=\;{ {\text{B}}^{2}}/{2 \mu _{0}} \times {\text{V}}$ $=\;\frac{10^{-14} \times 4 \times \pi ^{2} \times 4 \times 100}{2 \times 4 \pi \times 10^{-7}} \times (1 \times 10^{-9})$ $=2 \pi \times 10^{-14} {\text{J}}$ $=6.28 \times 10^{-14} {\text{J}}$

The magnetic field at the centroid of an equilateral triangle of side $a$ is given by $B = 3 \times \frac{\mu_0 I}{4\pi r} (\sin 60^\circ + \sin 60^\circ)$, where $r = \frac{a}{2\sqrt{3}}$ is the perpendicular distance from the centroid to any side.
Substituting $r = \frac{4\sqrt{3} \times 10^{-2}}{2\sqrt{3}} = 2 \times 10^{-2} \, \text{m}$ and $\sin 60^\circ = \frac{\sqrt{3}}{2}$:
$B = 3 \times \frac{\mu_0 I}{4\pi (2 \times 10^{-2})} (\sqrt{3}) = \frac{3\sqrt{3} \mu_0 I}{4\pi (2 \times 10^{-2})}$
Given $I = 2 \, \text{A}$ and $\mu_0 = 4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A}$, the expression becomes:
$B = \frac{3\sqrt{3} (4\pi \times 10^{-7}) (2)}{4\pi (2 \times 10^{-2})} = 3\sqrt{3} \times 10^{-5} \, \text{T}$
Comparing this to $B = \alpha \times 10^{-5} \, \text{T}$, we find $\alpha = 3\sqrt{3}$.

${\text{B}}=\;{\mu _{0} {\text{I}}}/{2 \pi (4) \times 10^{-2}} \times 2$ $=10^{-7} \times 30 \times 10^{2}$ $=30 \times 10^{-5} {\text{T}}$ $=300 \mu {\text{T}}$

$\vec{F} =q(\vec{v} \times \vec{B})$ $=10^{-6}({i}^{∧}-2{j}^{∧}+3{k}^{∧})\times (2{i}^{∧}+3{j}^{∧}-5{k}^{∧})$ $=10^{-6}(3{k}^{∧}+5{j}^{∧}+4{k}^{∧}+10{i}^{∧}+6{j}^{∧}-9{i}^{∧})$ $=10^{-6}({i}^{∧}+11{j}^{∧}+7{k}^{∧})$ $={\sqrt{171}}\times 10^{-6}$

Inside an ideal solenoid, the magnetic field $\vec{B}$ is directed along its axis. Since the particle moves along the same axis, its velocity vector $\vec{v}$ is parallel to $\vec{B}$ throughout its motion. The magnetic force on a charge $Q$ is given by $\vec{F_B} = Q(\vec{v} \times \vec{B})$. As $\vec{v} \parallel \vec{B}$, the cross product $\vec{v} \times \vec{B} = 0$, thus $\vec{F_B} = 0$. With no magnetic force, the only force acting on the particle is gravity, $F_g = mg$. Applying Newton's second law, $ma = mg$, yielding $a = g$.

The magnetic field at the centre of the loop is $B_c = \frac{\mu_0 I}{2R} = 16 \text{ } \mu\text{T}$.
The magnetic field on the axis at distance $x$ is $B_x = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$.
Substituting $x = \sqrt{3}R$ into the denominator gives $(R^2 + (\sqrt{3}R)^2)^{3/2} = (R^2 + 3R^2)^{3/2} = (4R^2)^{3/2} = 8R^3$.
The expression simplifies to $B_x = \frac{\mu_0 I R^2}{2(8R^3)} = \frac{\mu_0 I}{16R}$.
Comparing this to $B_c$, we find $B_x = \frac{1}{8} B_c = \frac{16 \text{ } \mu\text{T}}{8} = 2 \text{ } \mu\text{T}$.

$\Delta {\text{K}}= {\text{W}}_{ {\text{e}}}+ {\text{W}}_{ {\text{m}}}= {\text{W}}_{ {\text{e}}}+0$ $=∫_{0}^{ {\text{L}}} {\text{qEdx}}= {\text{qE}}_{0} ∫_{0}^{ {\text{L}}} {\text{e}}^{-\lambda {\text{x}}} {\text{dx}}$ $=\;{ {\text{qE}}_{0}}/{\lambda } [1- {\text{e}}^{-\lambda {\text{L}}}]$

${ \vec{\text{F}}}= {\text{q}} { \vec{\text{E}}}+ {\text{q}}({ \vec{\text{V}}} \times { \vec{\text{B}}})$ $=10^{-9} [0.4 { {\text{j}}}^{∧}+ ( {\text{v}}_{ {\text{x}}} { {\text{i}}}^{∧}+ {\text{v}}_{ {\text{y}}} { {\text{j}}}^{∧}) \times (4 \times 10^{-3}) { {\text{k}}}^{∧}]$ $(4 {i}^{∧}+2 {j}^{∧}) \times 10^{-10}$ $=10^{-9} (0.4 { {\text{j}}}^{∧}-4 \times 10^{-3} {\text{v}}_{ {\text{x}}} { {\text{j}}}^{∧}+4 \times 10^{-3} {\text{v}}_{ {\text{y}}} { {\text{i}}}^{∧})$On comparing${\text{v}}_{ {\text{y}}}=100, {\text{v}}_{ {\text{x}}}=50$ ${ \vec{\text{V}}}=50 { {\text{i}}}^{∧}+100 { {\text{j}}}^{∧}$

${\text{i}}_{ {\text{g}}} {\text{R}}_{ {\text{g}}}=2 (\;\frac{1}{2}- {\text{ig}})$ ${\text{i}}_{ {\text{g}}}=\;{1}/{ {\text{R}}_{ {\text{g}}}+2}$ $10= ( {\text{R}}_{ {\text{g}}}+470) {\text{i}}_{ {\text{g}}}$ $10=\;\frac{R_{g}+470}{R_{g}+2}$ $10 R_{ {\text{g}}}+20=R_{ {\text{g}}}+470$ ${\text{R}}_{ {\text{g}}}=50 Ω$

$\vec{\tau }={ \vec{\text{M}}} \times { \vec{\text{B}}}$ \Rightarrow 100 {\sqrt{2}} \times 3.14 \times 4 \times 10^{-4} \[ (;{{ {\text{k}}}^{∧}+{ {\text{i}}}^{∧}}/{{\sqrt{2}}}) ×(3 { {\text{i}}}^{∧}+2 { {\text{k}}}^{∧})]$ {\text{B}}_{0}$$\Rightarrow 16 \times 3.14 \times $[+3 { {\text{j}}}^{∧}-2 { {\text{j}}}^{∧}]$ \times 10^{-5}$$\Rightarrow 50.24 \times 10^{-5} { {\text{j}}}^{∧}$

Using Ampere's Law, $\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{\text{enclosed}}$, for a cylindrical conductor of radius $R$ carrying current $I$ uniformly, the enclosed current at a radial distance $r \leq R$ is $I_{\text{enclosed}} = I \frac{\pi r^2}{\pi R^2} = I \frac{r^2}{R^2}$. Integrating around a circular path of radius $r$ gives $B(2\pi r) = \mu_0 I \frac{r^2}{R^2}$, which simplifies to $B = \frac{\mu_0 I r}{2\pi R^2}$. At the axis of the conductor where $r = 0$, the magnetic field $B = 0$. Since $B$ increases linearly with $r$ inside ($0 < r < R$) and decreases as $\frac{1}{r}$ outside ($r > R$), the value $B=0$ at the axis is the minimum possible magnitude.

${\text{dM}}=( {\text{dI}}) \pi {\text{r}}^{2}=( {\text{IdN}}) \pi {\text{r}}^{2}$ $M=∫_{i}^{f} I (\;\frac{N}{r_{e x}-r_{i}}) d r \pi r^{2}$ {\text{M}}=\;{ {\text{I}} \pi {\text{N}}}/{ {\text{r}}_{ {\text{ex}}}- {\text{r}}_{ {\text{i}}}} \[;{ {\text{r}}^{3}}/{3}]$ { {\text{ri}}}^{ {\text{r}}{ {\text{ex}}}}$${\text{M}}=;{ {\text{I}} \pi {\text{N}}}/{3 ( {\text{r}}{ {\text{ex}}}- {\text{r}}{ {\in }}) } ( {\text{r}}{ {\text{ex}}}{ }^{3}- {\text{r}}{ {\text{i}}}{ }^{3})$${\text{M}}=;\frac{20 \times 10^{-3} \times 3.14 \times 200 $[6^{3}-3^{3}]$ 10^{-6}}{3[3] \times 10^{-2}}$${\text{M}}=;\frac{2373.84 \times 10^{-6}}{9 \times 10^{-2}}$${\text{M}}=263.76 \times 10^{-4}$${\text{M}}=2.64 \times 10^{-2} {\text{Am}}^{2}$

The force on a current-carrying wire of length $L$ due to another parallel wire is given by $F = \frac{\mu_0 I_1 I_2 L}{2\pi r}$.
Since current $I_P$ ($3\text{ A}$, upward) and $I_Q$ ($1\text{ A}$, downward) flow in opposite directions, the force $F_{PQ}$ is repulsive, pushing $Q$ towards $R$:
$F_{PQ} = \frac{(4\pi \times 10^{-7})(3)(1)(0.15)}{2\pi (0.03)} = 3 \times 10^{-6} \text{ N (towards } R)$
Since currents $I_Q$ ($1\text{ A}$, downward) and $I_R$ ($2\text{ A}$, downward) flow in the same direction, the force $F_{RQ}$ is attractive, pulling $Q$ towards $R$:
$F_{RQ} = \frac{(4\pi \times 10^{-7})(2)(1)(0.15)}{2\pi (0.02)} = 3 \times 10^{-6} \text{ N (towards } R)$
The net force on $Q$ is the sum of these two forces:
$F_{net} = 3 \times 10^{-6} + 3 \times 10^{-6} = 6 \times 10^{-6} \text{ N (towards } R)$

The magnetic field on the axis of a circular loop of radius $R$ at a distance $x$ from its center is given by $B = \frac{\mu_0 I R^2}{2(R^{2+}x^2)^{3/2}}$.

For loop $P$ with current $I$ and distance $r$, $B_P = \frac{\mu_0 I r^2}{2(r^{2+}r^2)^{3/2}} = \frac{\mu_0 I}{4\sqrt{2} r}$, directed towards $P$.
For loop $Q$ with current $4I$ and distance $r$, $B_Q = \frac{\mu_0 (4I) r^2}{2(r^{2+}r^2)^{3/2}} = \frac{4\mu_0 I}{4\sqrt{2} r}$, directed towards $Q$.
Since the currents are clockwise as viewed from $O$, $B_P$ and $B_Q$ point in opposite directions away from $O$.
The net field is $B_{net} = B_Q - B_P = \frac{4\mu_0 I - \mu_0 I}{4\sqrt{2} r} = \frac{3\mu_0 I}{4\sqrt{2} r}$ towards $Q$.

$\vec{F}_{B} \propto (\vec{V} \times \vec{B}) \;\; \Rightarrow \vec{F}_{B} ⟂ \vec{B}$ $\vec{a} ⟂ \vec{B} \longrightarrow \vec{a} \cdot \vec{B}=0$ $12- {\text{x}}=0 \;\; \Rightarrow {\text{x}}=12$

Magnetic field inside a solenoid$B\;=\mu _0 i (\;\frac{N}{L})$ $L\;=\;\frac{\mu _0 N i}{B}=\;\frac{4 \pi \times 10^{-7} \times 200 \times 0.29}{2.9 \times 10^{-4}}$ $\;=8 \pi cm$