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Magnetic flux phi(B) through a plane of ...

Magnetic flux `phi_(B)` through a plane of area 'A' placed in a uniform magnetic field `vceB` is given as:
`phi_(B)= vceB.vceA = BA cos theta`

where `theta` is the angle between `vceB` and `vceA`. The above relatio can be extended to curved surfaces and non-uniform fields. In general, if the magnetic field has different magnitudes an directions at various parts of a surface, then the magnetic flux through the surface is given by
`phi_(B)=int vec B.dvecA`
On the basis of the experimental observations, Faraday concluded that an emf is induced in a coil when magnetic flux through the coil changes with time. Faraday stated his conclusions in the form of a law called Faraday's law of electromagnetic induction. As per this law, the induced emf is given by
`varepsilon = -(dphi_(B))/dt`
The negative sign in the expression indicates the direction of induced emfe and hence the direction of current in a closed loop.
In the case of a closely would coil of N turns, change of flux associated with each turns is the same and so that total induced emf is given by `varepsilon=-N(dphi_(B))/dt`
From above relations it is clear that the magnetic flux can be changed by changing any one or more of the terms `vceB, vceA and theta`.
(a) Give SI unit of magnetic flux.
(b) How is it related to tesla ?
(c) Obtain dimensional formula of magnetic flux.
(d) A loop of area `4 xx 10^(-3)m^(2)` is placed with its plane perpendicular to a uniform magnetic field of 0.02 T. If the loop is quickly removed from the magnetic field within a time of 2 ms, what is the magnitude of induced emf across the two ends of the loop ?
(e) If the resistance of the loop be `0.2Omega` and a sensitive milliammeter be connected between the two ends of loop, what will be the reading of milliammeter?

Text Solution

Verified by Experts

(a) SI unit of magnetic flux is weber (Wb).
(b) `1Wb=1Tm^(2)`,
(c) Since `|varepsilon|=phi_(B)/t, "hence" [phi_(B)]=[varepsilon]xx[t]=[ML^(2)T^(-3)A^(-1)]xx[T]=[ML^(-2)T^(-2)A^(-1)]`
(d) Here `phi_(i) = BA = 0.02 xx 4 xx 10^(-3) = 8 xx 10^(-5) "Wb and" phi_(f)=0`
`therefore |varepsilon|=(phi_(i)-phi_(f))/t=(8 xx 10^(-5) -0)/(2xx10^(-3))= 4 xx 10^(-2)V=0.04V` (e) Induced current `I = (|varepsilon|)/R=(0.04)/(0.2)=0.2A or 200mA`
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Knowledge Check

  • If the magnetic field is parallel to a surface then, the magnetic flux through the surface is

    A
    zero
    B
    small but not zero
    C
    infinite
    D
    large but not infinite
  • A uniform magnetic field is represented by

    A
    parallel and equidistant field lines
    B
    non parallel but equidistant field lines
    C
    perpendicular lines
    D
    curved lines
  • A uniform magnetic field is represented by -

    A
    closed curves
    B
    parallel lines
    C
    convergent lines
    D
    divergent lines
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