","comment":{"@type":"Comment","text":"Force on moving particle in uniform perpendicular field is $f=q\\left( \\vec{V}\\times \\vec{B} \\right)$ and radius of circular motion is $r=\\dfrac{mv}{2B}$. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\left( 3 \\right)$$ and $$\\left( 1 \\right)$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\left( 4 \\right)$$ and $$\\left( 1 \\right)$$","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\left( 4 \\right)$$ and $$\\left( 2 \\right)$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\left( 3 \\right)$$ and $$\\left( 2 \\right)$$","position":0,"answerExplanation":{"@type":"Comment","text":" As for circular motion in a magnetic field $$B$$, radius $$R$$ is $$ \\\\R=\\dfrac{mV}{Bq}=\\dfrac{m}{q}\\dfrac{V}{B}\\,\\,\\,\\,\\,...\\left( 1 \\right) \\\\ $$ Force in a magnetic field is $$ \\\\ \\vec{F}=q\\left( \\vec{V}\\times \\vec{B} \\right) \\\\ $$ Since electron has negative charge, using the cross product rule, $$\\left( 4 \\right)$$ must be an electron, $$\\left( 1 \\right)$$ and $$\\left( 2 \\right)$$ can be proton or $$\\alpha $$-particle each and charge on neutron is zero, so $$\\left( 3 \\right)$$ must be neutron. $$ \\\\ \\text{From }\\left( 1 \\right)\\text{ }R \\propto m \\\\R \\propto \\dfrac{1}{q} \\\\ \\Rightarrow{{M}_{\\alpha }}=4{{M}_{P}} \\\\ \\Rightarrow{{Q}_{\\alpha }}=2{{Q}_{P}} \\\\ \\Rightarrow\\dfrac{{{M}_{\\alpha }}}{{{Q}_{\\alpha }}}=\\dfrac{4{{M}_{P}}}{{{Q}_{P}}} \\\\ \\text{Thus R}\\alpha >{{R}_{P}} \\\\ \\text{So, }\\left( 1 \\right)\\text{ is proton, }\\left( 2 \\right)\\text{ is}\\,\\alpha \\text{-particle}\\text{.}$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Electromagnetism Quiz 1","text":"A charge particle with charge $$Q$$ enters in a uniform magnetic field perpendicular to field with different speeds as $${{V}_{0}},2{{V}_{0}},3{{V}_{0}},4{{V}_{0}}$$. If the dimension or region of the magnetic field is $$\\left( b\\times b \\right){{m}^{2}}$$ then the maximum time spent by a charged particle in the field region with velocity. ","comment":{"@type":"Comment","text":" Time period of circular motion in the presence of a perpendicular magnetic field is $T=\\dfrac{2\\pi mv}{qB}$ use this and analyze the problem. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $${{V}_{0}}$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$2{{V}_{0}}$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$3{{V}_{0}}$$","position":2}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" None of above ","position":3,"answerExplanation":{"@type":"Comment","text":" In the uniform magnetic field $$ \\\\T=\\dfrac{2\\pi m}{qB}\\\\ $$ So time period is independent of radius of circle and velocity of particle i.e, time spent is independent of velocity. Hence, the option $$\\left( \\text{d} \\right)$$ is correct.","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Electromagnetism Quiz 1","text":" A charge particle is moving in a magnetic field. Motion of particles can be in the form of helix if. ","comment":{"@type":"Comment","text":"Use the formula, $\\vec{f}=q\\left( \\vec{V}\\times \\vec{B} \\right)$ and analyze the magnetic force value for different angles. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" angle between $$\\vec{V}$$ and $$\\vec{B}$$ is zero","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" charge particle moves in straight line ","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" angle should be a multiple of $$\\dfrac{\\pi }{2}$$","position":2}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" angle should not be a multiple of $$\\dfrac{\\pi }{2}$$","position":3,"answerExplanation":{"@type":"Comment","text":" $$ \\\\ $$ To move in a helix, A particle should have two perpendicular velocity components. $$ \\\\ $$ This can happen if a charge enters a magnetic field with an angle which is not an integral multiple of $$\\dfrac{\\pi }{2}$$. $$ \\\\ $$ Here $$V \\cos \\theta$$ and magnetic field $$B$$ is along the x-axis. $$ \\\\$$ While $$V\\sin \\theta$$ is along the y-axis. $$ \\\\ $$ $$V \\cos\\theta$$ will provide the forward velocity while $$V\\sin \\theta$$ will move the charge in circle due to $${{{\\vec{F}}}_{B}}$$ force perpendicular to it as $$ \\\\{{{\\vec{F}}}_{B}}=q\\left( \\vec{V}\\times \\vec{B} \\right) \\\\ $$ Together,they will make the charge move in helix. ","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Electromagnetism Quiz 1","text":"In a space, region $$\\left( 1 \\right)$$ is separated by region $$\\left( 2 \\right)$$ through an imaginary boundary $$XX'.$$ If region $$2$$ has magnetic field $$B={{B}_{o}}\\vec{K},$$ then a object of charge $$Q$$ enters the magnetic field at $$XX'$$ and leaves it in $$4$$ seconds. The object enters the field in direction $$\\hat{i}$$ will $$100J$$ of kinetic energy. Find the velocity of it’s entering region $$\\left( 2 \\right)$$, provided $${{B}_{o}}=2\\pi $$ Tesla and $$Q=1C$$.
","comment":{"@type":"Comment","text":"Its velocity is perpendicular to the magnetic field, it takes a circular motion for the charge. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$10\\text{m/s}$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$50\\text{m/s}$$","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" $$20\\text{m/s}$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$5\\text{m/s}$$","position":1,"answerExplanation":{"@type":"Comment","text":"
$$ \\\\ \\text{As V is along }\\hat{i},\\text{ it is perpendicular to }\\vec{B}={{B}_{o}}\\hat{K}. \\\\ \\text{So it will take a circular path due to magnetic force }{{{\\vec{F}}}_{B}}\\text{ acting as centripetal force}\\text{.} \\\\{{{\\vec{F}}}_{B}}=Q\\left( \\vec{V}\\times \\vec{B} \\right) \\\\ \\text{Let }V\\text{ be the velocity and }m\\text{ mass of object}\\text{.} \\\\ \\text{The total path will be a semicircle}\\text{.} \\\\ \\text{Time period for circular path in magnetic field }T=\\dfrac{2\\pi m}{Bq} \\\\ \\text{Time for semicircular path}=\\dfrac{T}{2}=\\dfrac{\\pi m}{Bq} \\\\ \\text{given }\\dfrac{\\pi m}{Bq}=4 \\\\ \\Rightarrow \\dfrac{\\pi \\times m}{\\left( 2\\pi \\right)\\times \\left( 1 \\right)}=4 \\\\ \\Rightarrow m=8kg \\\\ \\text{Given}\\dfrac{1}{2}m{{V}^{2}}=100 \\\\ \\Rightarrow \\dfrac{1}{2}\\times 8\\times {{V}^{2}}=100 \\\\ \\therefore V=5\\text{m/s}$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Electromagnetism Quiz 1","text":" A disc is rotating with angular velocity $$''W''$$ and charge $$''Q''$$ is uniformly spread over the area. Magnetic moment of the disc is. ","comment":{"@type":"Comment","text":" Magnetic moment is $m=i.A$. $'i'$ in rotating discs have charge $=\\dfrac{Q}{2\\pi }w$."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{QWR}{2}$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{QW{{R}^{2}}}{2}$$","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{QWR}{4}$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{QW{{R}^{2}}}{4}$$","position":1,"answerExplanation":{"@type":"Comment","text":"
$$ \\\\ \\text{Consider a thin circle of width }dx\\text{ and radius }'x'\\text{.} \\\\ \\text{Area of this strip}\\,\\text{ }dA'=\\left( 2\\pi x \\right)\\times dx \\\\ \\text{small amount of charge on this strip}=dQ \\\\dQ=\\dfrac{Q}{d{{R}^{2}}}\\times dA \\\\ \\Rightarrow dQ=\\dfrac{Q}{{{R}^{2}}}2xdx \\\\ \\text{As this strip moves with angular velocity }'W' \\\\ \\text{The small magnetic moment }dM\\text{ is} \\\\dM=dI\\left( \\pi {{x}^{2}} \\right) \\\\ \\Rightarrow dM=\\dfrac{WdQ}{2\\pi }\\left( \\pi {{x}^{2}} \\right) \\\\ \\Rightarrow dM=\\dfrac{W}{2}\\left( \\dfrac{Q}{{{R}_{3}}}2xdx \\right){{x}^{2}} \\\\ \\Rightarrow dM=\\dfrac{WQ}{{{R}^{2}}}{{x}^{3}}dx \\\\ \\text{Total magnetic moment from }x=o \\\\ \\text{ }x=R \\\\ \\Rightarrow M=\\int\\limits_{0}^{R}{dM=}\\int\\limits_{0}^{R}{\\dfrac{WQ}{{{R}^{2}}}{{x}^{3}}dx} \\\\ \\Rightarrow M=\\dfrac{WQ}{{{R}^{2}}}\\left[ \\dfrac{{{x}^{4}}}{4} \\right]_{0}^{R} \\\\ \\therefore M=\\dfrac{WQ{{R}^{2}}}{4}$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]}]}