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Paragraph for Question Nos. 12 and 14A thin non-conducting ring of mass m, radius a carrying a charge q can rotate freely about its own axis which is vertical. At the initial moment the ring was at rest in horizontal position and no magnetic field was present. At instant t = 0 A uniform magnetic field is switched on which is vertically downward and increases with time according to the law B = B0t. Neglecting magnetism induced due to rotational motion of ring.The magnitude of induced emf will bea)π a2 B0b)2 a2 B0c)zerod)1/2πa2 B0Correct answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Paragraph for Question Nos. 12 and 14A thin non-conducting ring of mass m, radius a carrying a charge q can rotate freely about its own axis which is vertical. At the initial moment the ring was at rest in horizontal position and no magnetic field was present. At instant t = 0 A uniform magnetic field is switched on which is vertically downward and increases with time according to the law B = B0t. Neglecting magnetism induced due to rotational motion of ring.The magnitude of induced emf will bea)π a2 B0b)2 a2 B0c)zerod)1/2πa2 B0Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Paragraph for Question Nos. 12 and 14A thin non-conducting ring of mass m, radius a carrying a charge q can rotate freely about its own axis which is vertical. At the initial moment the ring was at rest in horizontal position and no magnetic field was present. At instant t = 0 A uniform magnetic field is switched on which is vertically downward and increases with time according to the law B = B0t. Neglecting magnetism induced due to rotational motion of ring.The magnitude of induced emf will bea)π a2 B0b)2 a2 B0c)zerod)1/2πa2 B0Correct answer is option 'A'. Can you explain this answer?.

Solutions for Paragraph for Question Nos. 12 and 14A thin non-conducting ring of mass m, radius a carrying a charge q can rotate freely about its own axis which is vertical. At the initial moment the ring was at rest in horizontal position and no magnetic field was present. At instant t = 0 A uniform magnetic field is switched on which is vertically downward and increases with time according to the law B = B0t. Neglecting magnetism induced due to rotational motion of ring.The magnitude of induced emf will bea)π a2 B0b)2 a2 B0c)zerod)1/2πa2 B0Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.

Here you can find the meaning of Paragraph for Question Nos. 12 and 14A thin non-conducting ring of mass m, radius a carrying a charge q can rotate freely about its own axis which is vertical. At the initial moment the ring was at rest in horizontal position and no magnetic field was present. At instant t = 0 A uniform magnetic field is switched on which is vertically downward and increases with time according to the law B = B0t. Neglecting magnetism induced due to rotational motion of ring.The magnitude of induced emf will bea)π a2 B0b)2 a2 B0c)zerod)1/2πa2 B0Correct answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of Paragraph for Question Nos. 12 and 14A thin non-conducting ring of mass m, radius a carrying a charge q can rotate freely about its own axis which is vertical. At the initial moment the ring was at rest in horizontal position and no magnetic field was present. At instant t = 0 A uniform magnetic field is switched on which is vertically downward and increases with time according to the law B = B0t. Neglecting magnetism induced due to rotational motion of ring.The magnitude of induced emf will bea)π a2 B0b)2 a2 B0c)zerod)1/2πa2 B0Correct answer is option 'A'. Can you explain this answer?, a detailed solution for Paragraph for Question Nos. 12 and 14A thin non-conducting ring of mass m, radius a carrying a charge q can rotate freely about its own axis which is vertical. At the initial moment the ring was at rest in horizontal position and no magnetic field was present. At instant t = 0 A uniform magnetic field is switched on which is vertically downward and increases with time according to the law B = B0t. Neglecting magnetism induced due to rotational motion of ring.The magnitude of induced emf will bea)π a2 B0b)2 a2 B0c)zerod)1/2πa2 B0Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of Paragraph for Question Nos. 12 and 14A thin non-conducting ring of mass m, radius a carrying a charge q can rotate freely about its own axis which is vertical. At the initial moment the ring was at rest in horizontal position and no magnetic field was present. At instant t = 0 A uniform magnetic field is switched on which is vertically downward and increases with time according to the law B = B0t. Neglecting magnetism induced due to rotational motion of ring.The magnitude of induced emf will bea)π a2 B0b)2 a2 B0c)zerod)1/2πa2 B0Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice Paragraph for Question Nos. 12 and 14A thin non-conducting ring of mass m, radius a carrying a charge q can rotate freely about its own axis which is vertical. At the initial moment the ring was at rest in horizontal position and no magnetic field was present. At instant t = 0 A uniform magnetic field is switched on which is vertically downward and increases with time according to the law B = B0t. Neglecting magnetism induced due to rotational motion of ring.The magnitude of induced emf will bea)π a2 B0b)2 a2 B0c)zerod)1/2πa2 B0Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice JEE tests.