A rectangular loop has a sliding connector PQ of length l and resistance R\: \Omega and it is moving with a speed \upsilon as shown. The set-up is placed in a uniform magnetic field going into the plane of the paper. The three currents I_{1},I_{2}\: and \: I_{3} are

 

  • Option 1)

    I_{1}=I_{2}= \frac{Bl\upsilon }{6R},I= \frac{Bl\upsilon }{3R}

  • Option 2)

    I_{1}=-I_{2}= \frac{Bl\upsilon }{R},I= \frac{2Bl\upsilon }{R}

  • Option 3)

    I_{1}=I_{2}= \frac{Bl\upsilon }{3R},I= \frac{2Bl\upsilon }{3R}

  • Option 4)

    I_{1}=I_{2}= I= \frac{Bl\upsilon }{R},

 

Answers (1)

As we learnt in

Motional EMF -

\varepsilon = Blv
 

- wherein

B\rightarrow magnetic field

l\rightarrow length

v\rightarrow velocity perpendicular to uniform magnetic field.

 

 I={}\frac{\xi{}3R}{2}=\frac{2\xi }{3R}=\frac{2Bl\vartheta }{3R}

I_{1}=I_{2}=\frac{I}{2}=\frac{Bl\vartheta }{3R}


Option 1)

I_{1}=I_{2}= \frac{Bl\upsilon }{6R},I= \frac{Bl\upsilon }{3R}

This is incorrect option

Option 2)

I_{1}=-I_{2}= \frac{Bl\upsilon }{R},I= \frac{2Bl\upsilon }{R}

This incorrect option

Option 3)

I_{1}=I_{2}= \frac{Bl\upsilon }{3R},I= \frac{2Bl\upsilon }{3R}

This is correct option

Option 4)

I_{1}=I_{2}= I= \frac{Bl\upsilon }{R},

This is incorrect option

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