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

Here, current is uniformly distributed across the cross-section of the wire, therefore, current enclosed in the ampere-an path formed at a distance $r_1 (=\;\frac{a}{2})$ $= (\;\frac{\pi r_1^2}{\pi a^2}) \times I$, where $I$ is total current $\therefore$ Magnetic field at $P_1$ is $B_1=\;\frac{\mu _0 \times \;\text{ current enclosed }\;}{\;\text{ path }\;}$ $\Rightarrow B_1=\;\frac{\mu _0 \times (\frac{\pi r_1^2}{\pi a^2}) \times I}{2 \pi r_1}$ $=\;\frac{\mu _0 \times I r_1}{2 \pi a^2}$ Now, magnetic field at point $P_2$,$B_2=\;\frac{\mu _0}{2 \pi } \cdot \;\frac{I}{(2 a)}=\;\frac{\mu _0 I}{4 \pi a}$ $\therefore$ Required ratio $=\;\frac{B_1}{B_2}=\;\frac{\mu _0 I r_1}{2 \pi a^2} \times \;\frac{4 \pi a}{\mu _0 I}$ $=\;\frac{2 r_1}{a}=\;\frac{2 \times \frac{a}{2}}{a}=1$.

Here, $\vec{E}$ and $\vec{B}$ are perpendicular to each other and the velocity $\vec{v}$ does not change; therefore$q E=q v B \Rightarrow v=\;\frac{E}{B}$Also, ${|\;\frac{\vec{E} \times \vec{B}}{B^2}|} =\;\frac{E B sin \;\theta }{B^2}$ $=\;\frac{E B sin \;99^{\circ}}{B^2}=\;\frac{E}{B}={|\vec{v}|}=v$

NOTE : When a charged particle enters a magnetic field at a direction perpendicular to the direction of motion, the path of the motion is circular. In circular motion the direction of velocity changes at every point (the magnitude remains constant).Therefore, the tangential momentum will change at every point. But kinetic energy will remain constant as it is given by $\;\frac{1}{2} m v^2$ and $v^2$ is the square of the magnitude of velocity which does not change.

The charged particle will move along the lines of electric field (and magnetic field). Magnetic field will exert no force. The force by electric field will be along the lines of uniform electric field. Hence the particle will move in a straight line.

$\;\frac{B_2}{B_1}=\;\frac{\mu _0 n_2 i_2}{\mu _0 n_1 i_1}$ $\Rightarrow \;\frac{B_2}{6.28 \times 10^{-2}}=\;\frac{100 \times \frac{i}{3}}{200 \times i}$ $\Rightarrow B_2=\;\frac{6.28 \times 10^{-2}}{6}$ $=1.05 \times 10^{-2} {\text{Wb}} / {\text{m}}^2$

$\;\frac{F}{ℓ}=\;\frac{\mu _0 i_1}{2 \pi d}=\;\frac{\mu _0 i^2}{2 \pi d}$(attractive as current is in the same direction)

Equating magnetic force to centripetal force.$\;\frac{m V^2}{r}=q v B sin \;90^{\circ}$Time to complete one revolution.$T=\;\frac{2 \pi r}{v}=\;\frac{2 \pi m}{q B}$

Due to electric field, it experiences force and decelerates i.e. its velocity decreases.

The magnetic field due to circular coil 1 and 2 are$\;B_1=\;\frac{\mu _0 i_1}{2 r}=\;\frac{\mu _0 i_1}{2 (2 \pi \times 10^{-2}) }$ $\;=\;\frac{\mu _0 \times 3 \times 10^2}{4 \pi }$ $\;B_2=\;\frac{\mu _0 i_2}{2 (2 \pi \times 10^{-2}) }=\;\frac{\mu _0 \times 4 \times 10^2}{4 \pi }$ $\;B={\sqrt{B_1^{2+}B_2^2}}=\;\frac{\mu _0}{4 \pi } \cdot 5 \times 10^2$ $\;\Rightarrow B=10^{-7} \times 5 \times 10^2$ $\;\Rightarrow B=5 \times 10^{-5} {\text{Wb}} / {\text{m}}^2$

Using Ampere's law at a distance $r$ from axis, $B$ is same from symmetry.$∫ B \cdot d l=\mu _0 i$i.e., $B \times 2 \pi r=\mu _0 i$Here $i$ is zero, for $r< R$, whereas $R$ is the radius$\therefore B=0$

Force between two long conductor carrying current,$F=\;\frac{\mu _0}{4 \pi } \;\frac{2 I_1 I_2}{d} \times ℓ$ $F^{'}=-\;\frac{\mu _0}{4 \pi } \;\frac{2 (2 I_1) I_2}{3 d} ℓ$ $\therefore \;\frac{F^{'}}{F}=\;\frac{-2}{3}$

The magnetic field at a point on the axis of a circular loop at a distance $x$ from center is,$B=\;\frac{\mu _0 i a^2}{2 (x^{2+}a^2) 3 / 2} \;\; B^{'}=\;\frac{\mu _0 i}{2 a}$ $\therefore B^{'}=\;\frac{B \cdot (x^{2+}a^2) ^{3 / 2}}{a^3}$Put x=4 & a=3 \Rightarrow B^{'}=\;\frac{54 (5^3) }{3 \times 3 \times 3}=250 \mu T

KEY CONCEPT : Magnetic field at the center of a circular coil of radius $R$ carryingcurrent is $B=\;\frac{\mu _0 i}{2 R}$ Given: $n \times (2 \pi r^{'}) =2 \pi R$ $\Rightarrow n r^{'}=R$ $B^{'}=\;\frac{n \cdot \mu _0 i}{2 r^{'}}$from (1) and (2), $B^{'}=\;\frac{n \mu _0 i \cdot n}{2 \pi R}=n^2 B$

The work done, $d W=F d s \cos \theta$The angle between force and displacement is $90^{\circ}$.Therefore work done is zero.

The situation is shown in the figure.$F_E=$ Force due to electric field$F_B=$ Force due to magnetic fieldIt is given that the charged particle remains moving along $X$-axis (i.e. undeviated).Therefore $F_B=F_E$ $\;\Rightarrow q v B=q E$ $\;\Rightarrow B=\;\frac{E}{v}=\;\frac{10^4}{10}$ $\;=10^3 \;\text{ weber }\; {/} {\text{m}}^2$

Magnetic field due to current in wire 1 at point $P$ distant $r$ from the wire is$B=\;\frac{\mu _0}{4 \pi } \;\frac{i_1}{r}[\cos \theta +\cos \theta ]$ $B=\;\frac{\mu _0}{2 \pi } \;\frac{i_1 \cos \theta }{r}$ (directed perpendicular to the plane of paper, inwards) The force exerted due to this magnetic field on current element $i_2 d l$ is $d F=i_2 d l B \sin 90^{\circ}$ $\therefore d F=i_2 d l[\;\frac{\mu _0}{2 \pi } \;\frac{i_1 \cos \theta }{r}]$ $=\;\frac{\mu _0}{2 \pi r} i_1 i_2 d l \cos \theta$

KEY CONCEPT : We know that the magnetic field produced by a current carrying circular coil of radius $r$ at its center is $B=\;\frac{\mu _0}{4 \pi } \;\frac{I}{r} \times 2 \pi$ Here $B_A=\;\frac{\mu _0}{4 \pi } \;\frac{I}{R} \times 2 \pi$ and $B_B=\;\frac{\mu _0}{4 \pi } \;\frac{2 I}{2 R} \times 2 \pi$ $\Rightarrow \;\frac{B_A}{B_B}=1$

KEY CONCEPT : The time period of a charged particle $(m, q)$ moving in a magnetic field $(B)$ is $T=\;\frac{2 \pi m}{q B}$ The time period does not depend on the speed of the particle.

KEY CONCEPT : When a charged particle enters perpendicular to a magnetic field, then it moves in a circular path of radius. $r=\;\frac{p}{q B}$ where $q=$ Charge of the particle $p=$ Momentum of the particle $B=$ Magnetic field Here $p, q$ and $B$ are constant for electron and proton, therefore the radius will be same.

When current is passed through a spring then current flows parallel in the adjacent turns. NOTE : When two wires are placed parallel to each other and current flows in the same direction, the wires attract each other. Similarly here the various turns attract each other and the spring will compress.