A ring is rotated about diametric axis in a uniform magnetic field perpendicular to the plane of the ring. If initially the plane of the ring is perpendicular to the magnetic field. Find the instant of time at which EMF will be maximum & minimum respectively :

Question

Arun · Answer

ω = 2π/T = π/5 When ωt = π/2 ∵ ϕ will be minimum ∴ e will be maximum When ωt = π ∵ ϕ will have maximum ∵ e will be minimum

Vikas TU · Answer

ω = 2π/T = π/5 When ωt = π/2 ∵ ϕ will be minimum ∴ e will be maximum t = 2.5 sec wt = pi Phi will be max, e will be minimum t = 5 sec

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A ring is rotated about diametric axis in a uniform magnetic field perpendicular to the plane of the ring. If initially the plane of the ring is perpendicular to the magnetic field. Find the instant of time at which EMF will be maximum & minimum respectively :

A ring is rotated about diametric axis in a uniform magnetic field perpendicular to the plane of the ring. If initially the plane of the ring is perpendicular to the magnetic field. Find the instant of time at which EMF will be maximum & minimum respectively :

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A ring is rotated about diametric axis in a uniform magnetic field perpendicular to the plane of the ring. If initially the plane of the ring is perpendicular to the magnetic field. Find the instant of time at which EMF will be maximum & minimum respectively :

A ring is rotated about diametric axis in a uniform magnetic field perpendicular to the plane of the ring. If initially the plane of the ring is perpendicular to the magnetic field. Find the instant of time at which EMF will be maximum & minimum respectively :

2 Answers

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Other Related Questions on Magnetism

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