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  • The figure shows a rectangular conducting frame MNOP of resistance R placed partly in a perpendicular magnetic field and moved with velocity <img alt="\vec{v}" src=

The figure shows a rectangular conducting frame MNOP of resistance R placed partly in a perpendicular magnetic field \vec{B} and moved with velocity \vec{v} as shown in the figure.

Obtain the expressions for the

(a) force acting on the arm ‘ON’ and its direction, and

(b) power required to move the frame to get a steady emf induced between the arms MN and PO.

 

 
 
 
 
 

Answers (1)

(a) The induced emf in the moving conductor MNOP

           

        e=BlV

where B = Magnetic Field

             l= length of the conductor

The induced current, i=\frac{e}{r}=\frac{BlV}{r}

Where, r= radius

Hence, the force on the arm 'ON' (F)=Bil

                On putting the value of 'i' we have force,

                (F)=\frac{B\times B\times l\times V\times l}{r}

The force is directed in the direction opposite to Velocity of the rod (v).

(b) We know that,

   Power:P=F\times V

Where F = Force

 V = velocity
We know, F=\frac{B^{2}l^{2}V}{r}

    therefore, P=\frac{B^{2}l^{2}V^2}{r}

Posted by

Safeer PP

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