","comment":{"@type":"Comment","text":"Magnetic flux $\\phi =B.A,$ use the formula and analyze the problem. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" Case $$1$$ ","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" Same in both","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" All are correct","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" Case $$2$$","position":1,"answerExplanation":{"@type":"Comment","text":" $$\\text{As induced emf} \\\\V=\\dfrac{d\\phi }{dt} \\\\ \\Rightarrow V=-\\dfrac{d\\left( \\vec{B}.\\vec{A} \\right)}{dt} \\\\ \\Rightarrow V=-\\dfrac{d}{dt}\\left( BA\\cos \\theta \\right) \\\\ \\Rightarrow V=-B\\cos \\theta \\dfrac{dA}{dt} \\\\ \\text{In first case }{{\\text{B}}_{2}}>{{B}_{1}} \\\\ \\text{Thus }{{V}_{2}}>{{V}_{1}}$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Magnetic flux Quiz 1","text":" As viewed from the left, the current flows in an anti-clockwise direction. A battery attached to a solenoid of $$N$$ turn of $$L$$ length. Directions of magnetic field lines are: $$ \\\\ $$","comment":{"@type":"Comment","text":"Solenoid can act as a bar magnet, identifying the poles of magnet. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" Right to left","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" No field lines","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" Cannot say","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" From left to right","position":0,"answerExplanation":{"@type":"Comment","text":"
$$\\\\$$ If we look from the left, we can imagine seeing a circular loop with current flowing in anticlockwise direction. Thus it can be equated as a north pole of solenoid, with magnetic field lines emerging from the face. So the magnetic field lines move from right to left inside the coil.","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Magnetic flux Quiz 1","text":" A conducting coil rotating around an axis perpendicular to the centre. It is surrounded by a magnetic field. The direction of induced current will be $$ \\\\ $$
","comment":{"@type":"Comment","text":"For induction of fields there should be change in flux."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" Anti-clockwise","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" Clockwise","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" Current induced for fraction of time","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" No current induced","position":2,"answerExplanation":{"@type":"Comment","text":" $$\\text{As induced emt}=-\\dfrac{d\\phi }{dt} \\\\ \\Rightarrow -\\dfrac{d\\left( \\vec{B}.\\vec{A} \\right)}{dt} \\\\ \\Rightarrow -\\dfrac{d}{dt}\\left( BA\\cos \\theta \\right) \\\\ \\text{B,A and }\\cos \\theta \\text{ all are constant } \\\\ \\text{So Induced emf}=0 \\\\ \\text{Induced current}=0$$ ","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Magnetic flux Quiz 1","text":"The velocity of a rod moving in a magnetic field $$B=2\\hat{k}$$ Tesla is $$\\text{50 m/s}$$. If a resistance $$R$$ is connected across the rod, then find the total heat dissipated across the resistor till the rod stops moving. The mass of rod is $$\\text{10 kg}$$, the length is $$\\text{1 m}$$. $$ \\\\ $$
","comment":{"@type":"Comment","text":"Heat dissipated in the resistor is equal to the kinetic energy."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" Cannot say as given data is insufficient ","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$10000\\text{ J}$$","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" None of these","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$12500\\text{ J}$$","position":1,"answerExplanation":{"@type":"Comment","text":" The kinetic energy of the rod is dissipated as heat because motion the rod produces current in the resistance $$R$$. $$ \\\\ \\Rightarrow K.E=\\dfrac{1}{2}m{{v}^{2}} \\\\ \\Rightarrow K.E=\\dfrac{1}{2}\\times 10\\times {{\\left( 50 \\right)}^{2}} \\\\ \\therefore K.E=12500\\text{ J} \\\\ $$ The energy will be dissipated across the resistor since the magnetic field is conservative in nature.","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Magnetic flux Quiz 1","text":"If $${{\\phi }_{B}}=2t+\\dfrac{3}{\\pi }\\sin \\left( \\dfrac{\\pi }{3}t \\right)$$ where $$t$$ is time then magnitude of emf induced at $$t=1\\sec $$ is","comment":{"@type":"Comment","text":" $\\dfrac{d\\phi }{dt}=\\dfrac{{{\\phi }_{2}}-{{\\phi }_{1}}}{{{t}_{2}}-{{t}_{2}}}=emf$ induced in the circuit, find flux for time duration. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$105$$ volt","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{2+\\sqrt{3}}{2}$$volt","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{2-\\sqrt{3}}{2}$$volt","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$2.5$$volt","position":2,"answerExplanation":{"@type":"Comment","text":" $$emf=\\dfrac{d\\phi }{dt}=2+\\dfrac{3}{\\pi }\\times \\dfrac{\\pi }{3}\\cos \\left( \\dfrac{\\pi }{3}t \\right) \\\\ \\Rightarrow emf\\left( t \\right)=2+\\cos \\left( \\dfrac{\\pi }{3}t \\right) \\\\ \\text{At }t=1\\sec\\\\ \\Rightarrow emf =2+\\cos \\left( \\dfrac{\\pi }{3}\\right)\\\\ \\therefore emf=2.5\\,V$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]}]}