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  • Can someone help me with this, A uniform conducting rod of mass M and length l oscillates in a vertical plane about a fixed horizontal axis passing through its one end with angular amplitude θ. There exists a constant and uniform horizontal mag

A uniform conducting rod of mass M and length l oscillates in a vertical plane about a fixed horizontal axis passing through its one end with angular amplitude θ. There exists a constant and uniform horizontal magnetic field of induction B perpendicular to the plane of oscillation. The maximum e.m.f. induced in the rod is

  • Option 1)

    \frac{B}{8}\sqrt{27l^{3}g\left ( 1-cos\theta \right )}

  • Option 2)

    \frac{B}{8}\sqrt{27l^{3}g\left ( 1+cos\theta \right )}

  • Option 3)

    B\sqrt{\frac{3l^{3}g\left ( 1-cos\theta \right )}{4}}

  • Option 4)

    B\sqrt{\frac{3l^{3}g\left ( 1+cos\theta \right )}{4}}

 

Answers (1)

best_answer

As we discussed in concept

Motional E.m.f due to rotational motion -

Conducting rod \rightarrow

\varepsilon =\frac{1}{2}Bl^{2}\omega =Bl^{2}\pi\nu

 \nu \rightarrow f\! r\! equency

T \rightarrow Time\; period

 

- wherein

 

 Maximum \varepsilon mf induced when rod is vertical as it is in this position that its velocity is maximum.

From energy conservation

\frac{1}{2}IW^{2} = mg\: \frac{l}{2}(1-cos\theta)

or \frac{ml^{2}}{3}\:.w^{2}=mgl(1-cos\theta)

or w=\sqrt{\frac{3g}{l}\:.(1-cos\theta)}

\varepsilon mf \:Induced = \frac{BWl^{2}}{2}=\frac{B}{2}\:.\:l^{2}\:.\:\sqrt{\frac{3g}{l}\:(1-cos\theta)}

\xi \:=\:B\sqrt{\frac{3l^{3}g}{4}\:.\: (1-cos\theta)}


Option 1)

\frac{B}{8}\sqrt{27l^{3}g\left ( 1-cos\theta \right )}

This option is incorrect.

Option 2)

\frac{B}{8}\sqrt{27l^{3}g\left ( 1+cos\theta \right )}

This option is incorrect.

Option 3)

B\sqrt{\frac{3l^{3}g\left ( 1-cos\theta \right )}{4}}

This option is correct.

Option 4)

B\sqrt{\frac{3l^{3}g\left ( 1+cos\theta \right )}{4}}

This option is incorrect.

Posted by

Aadil

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