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  • (a) A metallic rod of length ‘l’ and resistance ‘R’ is rotated with a frequency ‘v’ with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius ‘l’, abou

(a) A metallic rod of length ‘l’ and resistance ‘R’ is rotated with a frequency ‘v’ with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius ‘l’, about an axis passing through the centre and perpendicular to the plane of the ring. A constant and uniform magnetic field ‘B’ parallel to the axis is present everywhere.

(i) Derive the expression for the induced emf and the current in the rod.

(ii) Due to the presence of current in the rod and of the magnetic field, find the expression for the magnitude and direction of the force acting on this rod.

(iii) Hence, obtain an expression for the power required to rotate the rod.

(b) A copper coil is taken out of a magnetic field with a fixed velocity. Will it be easy to remove it from the same field if its ohmic resistance is increased?

 

 

Answers (1)

(i) To determine the induced emf, 

      

a) As the rod is rotated, free electrons in the rod move towards the outer end due to Lorentz force and get distributed over the ring. Thus, the resulting separation of charges produces an emf across the ends of the rod. At a certain value of emf, there is no more flow of electrons and a steady-state is reached. The magnitude of the emf generated across a length dr of the rod is 

\\dE=Bvdr\\\\E=\int_{0}^{l}Bvdr=\int_{0}^{l}B\omega rdr=\frac{1}{2}Bl^2\omega

Now, we know that the induced current is given as ;

I=\frac{\varepsilon }{R}

I=\frac{Bl^{2}w}{2R}

(ii) The magnitude of the force is given as ;

F=BIl

we know that

, I=\frac{Bl^{2}w}{2R}

Hence,

F=B\times \frac{Bl^{2}w\times l}{2R}

F=\frac{B^{2}l^{3}w}{2R}

(iii) We know that, power 

P=I^{2}R

On putting the value of 'I', we get

P=\left ( \frac{Bl^{2}w}{2R} \right )^{2}R

P=\frac{B^{2}l^{4}w^{2}}{4R}

(b) the induced current decreases, so it is easy to remove a copper coil from the same field if its atomic resistance is increased.

Posted by

Safeer PP

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