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Using Faraday's law of electromagnetic induction, derive an equation for motional emf. |
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Answer» Solution :Consider a rectangular conducting loop of width 1 in a uniform magnetic field B which is perpendicular to the plane of the loop and is directed inwards. II. A part of the loop is in the magnetic field while the remainig par t is outside the loop as shown in Figure. When the loop is pulled with a constant velocity `vec(v)` to the right, the area of the portion of the loop within the magnetic field will decrease. iv. THUS, the flux linked with the loop will also decrease. According to Faraday's law, an electric current is induced in the loop which flows ina direction so as to oppose the pull of the loop. v. Let x be the length of the loop which is still within the magnetic field, then its area is lx. The magnetic flux linked with the loop is `phi_(B)=int_(A)B*dvec(A)=BAcostheta` Here`theta=0^(@)andcos0^(@)=1` = BA `phi_(B)=Blx` vi. As this magnetic flux DECREASES due to the movement of the loop, the magnitude ofthe induced emf is given by `EPSILON=(dPhi_(B))/(dt)=(d)/(dt)(Blx)` vii. Here, bothe B and l are constants. Therefore, `epsilon=Bl(dx)/(dt)` `epsilon=Blv` where `v=(dx)/(dt)` is the velocity of the loop. This emf is KNOWN as motional emf since it is produced due to the movement of the loop in the magnetic field. |
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