Electromagnetic Induction
- What is the self-inductance of a coil which produces 5V when the current changes from 3 ampere to 2 ampere in one millisecond?
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L = e di dt = e dt = 5 × 10-3 H = 5 mH di (3 - 2) Correct Option: B
L = e di dt = e dt = 5 × 10-3 H = 5 mH di (3 - 2)
- If N is the number of turns in a coil, the value of self inductance varies as
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L = Nφ ; φ = BA; i B = μ0Ni 2R ⇒ L = N = μ0Ni A i 2R = μ0N2 A ⇒ L ∝ N2 2R Correct Option: C
L = Nφ ; φ = BA; i B = μ0Ni 2R ⇒ L = N = μ0Ni A i 2R = μ0N2 A ⇒ L ∝ N2 2R
- A rectangular coil of 20 turns and area of cross-section 25 sq. cm has a resistance of 100Ω. If a magnetic field which is perpendicular to the plane of coil changes at a rate of 1000 teals per second, the current in the coil is
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i = e = nAdB/dt R R = 20 × (25 × 10-4) × 1000 = 0.5 A 100 Correct Option: C
i = e = nAdB/dt R R = 20 × (25 × 10-4) × 1000 = 0.5 A 100
- The total charge induced in a conducting loop when it is moved in a magnetic field depend on
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q = idt = 1 edt R = 1 - dφ dt R dt = 1 dφ R
(taking only magnitude of e)
Hence, total charge induced in the conducting loop depends upon the total change in magnetic flux.Correct Option: C
q = idt = 1 edt R = 1 - dφ dt R dt = 1 dφ R
(taking only magnitude of e)
Hence, total charge induced in the conducting loop depends upon the total change in magnetic flux.
- A 100 millinery coil carries a current of 1A. Energy stored in its magnetic field is
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E = 1 Li2 2 = 1 × (100 × 10-3) × 12 = 0.05 J 2 Correct Option: C
E = 1 Li2 2 = 1 × (100 × 10-3) × 12 = 0.05 J 2