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

The magnetic field produced by one side of an equilateral triangle of side $a$ at its center is given by $B_{side} = \frac{\mu_0 I}{4 \pi r} (\sin \theta_1 + \sin \theta_2)$, where $r = \frac{a}{2\sqrt{3}}$ is the perpendicular distance and $\theta_1 = \theta_2 = 60^\circ$.
Substituting $r$ and the angles: $B_{side} = \frac{\mu_0 I}{4 \pi (a/2\sqrt{3})} (2 \sin 60^\circ) = \frac{\mu_0 I \sqrt{3}}{2 \pi a / \sqrt{3}} = \frac{3 \mu_0 I}{2 \pi a}$.
For an equilateral triangle, the total field at the center is $B = 3 \times B_{side} = \frac{9 \mu_0 I}{2 \pi a}$.
Substituting the given values $\mu_0 = 4\pi \times 10^{-7} \, T \cdot m/A$, $I = 1 \, A$, and $a = 4.5 \times 10^{-2} \, m$:
$B = \frac{9 \times (4\pi \times 10^{-7}) \times 1}{2 \pi \times 4.5 \times 10^{-2}} = \frac{36\pi \times 10^{-7}}{9\pi \times 10^{-2}} = 4 \times 10^{-5} \, Wb/m^2$

$m=i\lambda a^2$ $m'=2m=2i\pi a^2=i\pi a^{+2}$ $a'=\sqrt2a$ $B_1/B_2=\frac{\mu _0i}{2a}/\frac{\mu _0i}{2a\sqrt2}=\sqrt2$

PE = -PE cos θτ = PE sin θ$\tau _\text{max}$ when $\theta =90°$PE = 0

$F_e=F_n=0$ $F_e=-F_m$ $=-q(\vec{v}\times \vec{B})$ $=-υ_0V_0[(3i^{\wedge }-j^{\wedge }+2k^{\wedge })\times (i^{\wedge }+2j^{\wedge }-4k^{\wedge })]$ $=-υ_0V_0(14i^{\wedge }+7k^{\wedge })$

The magnetic field $\vec{B}$ produced by a current $I$ flowing in the $+\hat{z}$ direction is given by $\vec{B} = B \hat{\theta}$ at a radial distance $r$. The magnetic force on a charge $q$ is $\vec{F} = q(\vec{v} \times \vec{B})$. Given the charge $q < 0$ and assuming a radial velocity component $\vec{v} = v_r \hat{r}$, the force becomes:
$\vec{F} = q(v_r \hat{r} \times B \hat{\theta}) = q v_r B \hat{k}$
Since $\vec{F}$ must be parallel to the current direction ($\hat{k}$), we have $\vec{F} = F \hat{k}$, where $F$ is a scalar. Substituting $q = -|q|$, we get $\vec{F} = -|q| v_r B \hat{k}$. For this force to be in the same direction as the current (positive $\hat{k}$), the radial velocity $v_r$ must be negative. A negative $v_r$ implies the charge is moving towards the wire.

$B_1\text{sin}30°=B_2\text{sin}45°$ $B_2=B_1/\sqrt2=10.60mT\approx 11mT$

B_A=\mu _0/{4\pi }{2\pi i}/{(l/{2\pi }) $B_B=[\mu _0/{4\pi }i/{(l/8)}(sin45°+sin45°)]\times 4$ $B_A/B_B=\pi ^2/{8\sqrt2}$

$i_1=V_E/{R+G}$ $i=V_E/{R+\frac{GS}{G+S}}$ $i_1/2G=(i-i_1/2)S$ $i_1(G+S)=2iS$substituting $i_1$ and i we get$S(R+G)=RG$

$tan\theta =\frac{Fl}{\lambda lg}={{(\frac{\mu _0I^2}{4\pi Lsin\theta })}l}/{\lambda lg}$ $\Rightarrow 1=2sin\theta \sqrt{\frac{\pi \lambda Lg}{\mu _0cos \theta }}$

Since $\vec{B}$ is uniform, only torque acts on a current carrying loop.$\tau ^{\to }=(I\vec{A})\times B$ $\vec{A}=Ak^{∧}$ for (b) and $\vec{A}=-Ak^{∧}$ for (d)$\therefore \tau ^{\to }=0$ for both these cases.The energy of the loop in the $\vec{B}$ field is: $U=-I\vec{A}·\vec{B}$, which is minimum for (b).

P=${\text{Work Done}}/\text{Time}={∫\text{Fdx}}/\text{t}={∫\text{IlbB.dx}}/\text{t}$ $=\frac{∫^2_0(10)(3)(3\times 10^{-4}e^{-0.2x})dx}{5\times 10^{-1}}$ \frac{9\times 10^{-3}}{5\times 10^{-3}}\[e^\frac{-0.2x}{-0.2}]^2_0=9$[1-e^{-0.4}]$=2.97w$

$r=\;{{\sqrt{2 m v}}}/{q B} \Rightarrow r \times v \;{{\sqrt{m}}}/{q}$Thus we have, $r_\alpha =r_p< r_d$

The magnetic field due a disc is given as$B=\;\frac{h_0 \omega Q}{2 \pi R} \;\text{ i.e., }\; B \propto \;\frac{1}{R}$

Current in a small element, $d l=\;\frac{d \theta }{\pi } I$Magnetic field due to the element$d B=\;\frac{\mu _0}{4 \pi } \;\frac{2 d l}{R}$The component $d B \cos \theta$, of the field is canceled by another opposite component.Therefore, $B_{n e t}=∫ d B sin \;\theta =\;\frac{\mu _0 I}{2 \pi ^2 R_0}$ $∫_{0}^{\pi } sin \;\theta d \theta =\;\frac{\mu _0 I}{\pi ^2 R}$

The magnetic field varies inversely with the distance for a long conductor. That is, $B \propto \;\frac{1}{d}$. According to the magnitude and direction shown graph (1) is the correct one.

$\vec{F}=I(\vec{ℓ} \times \vec{B})$The force on $A D$ and $B C$ due to current $I_1$ is zero. This is because the directions of current element $\vec{I d ℓ}$ and magnetic field $\vec{B}$ are parallel.

The magnetic field at $O$ due to current in $D A$ is $B_1=\;\frac{\mu _0}{4 \pi } \;\frac{I}{a} \times \;\frac{\pi }{6}$(directed vertically upwards) The magnetic field at $O$ due to current in $B C$ is $B_2=\;\frac{\mu _0}{4 \pi } \;\frac{I}{b} \times \;\frac{\pi }{6}$ (directed vertically downwards) The magnetic field due to current $A B$ and $C D$ at $O$ is zero.therefore the net magnetic field is $B=B_1-B_2$ (directed vertically upwards)$=\;\frac{\mu _0}{4 \pi } \;\frac{I}{a} \;\frac{\pi }{6}-\;\frac{\mu _0}{4 \pi } \;\frac{I}{b} \times \;\frac{\pi }{6}$ $=\;\frac{\mu _0 I}{24} (\;\frac{1}{a}-\;\frac{1}{b})$ $=\;\frac{\mu _0 I}{24 a b}(b-a)$

The magnetic field is$\;B=\;\frac{\mu _0}{4 \pi } \;\frac{2 I}{r}$ $\;=10^{-7} \times \;\frac{2 \times 100}{4}$ $\;=5 \times 10^{-6} T$According to right hand palm rule, the magnetic field is directed towards south.

Clearly, the magnetic fields at a point $P$, equidistant from $A O B$ and $C O D$ will have directions perpendicular to each other, as they are placed normal to each other. $\therefore$ Resultant field, $B={\sqrt{B_1^{2+}B_2^2}}$But $B_1=\;\frac{\mu _0 I_1}{2 \pi d}$ and $B_2=\;\frac{\mu _0 I_2}{2 \pi d}$ $\therefore B=\sqrt{ (\;\frac{\mu _0}{2 \pi d}) ^2 (I_1^{2+}I_2^2) }$or, $B=\;\frac{\mu _0}{2 \pi d} (I_1^{2+}I_2^2) ^{1 {/} 2}$

There is no current inside the pipe. Therefore$\;∮ \overline{B} \cdot \overline{d ℓ}=\mu _0 I$ $\;I=0$ $\; \therefore B=0$