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  • I need help with A metallic rod of mass per unit length 0.5 kg m-1 is lying horizontally on a smooth inclined plane which makes an angle of 30o with the horizontal. The rod is not allowed to slide down by flowing a current through it when a magnetic

A metallic rod of mass per unit length 0.5 kg m-1 is lying horizontally on a smooth inclined plane which makes an angle of 30o with the horizontal. The rod is not allowed to slide down by flowing a current through it when a magnetic field of induction 0.25 T is acting on it in the vertical direction. The current flowing in the rod to keep it stationary is

  • Option 1)

    14.76 A

  • Option 2)

    5.98 A

  • Option 3)

    7.14 A

  • Option 4)

    11.32 A

 

Answers (1)

best_answer

As we have learned

Force on a current carrying conductor -

underset{dF}{ ightarrow} =i (underset{dl}{ ightarrow} imes underset{B}{ ightarrow})

- wherein

idl=current element 

 

To keep the rod stationary 

F= i\hat{l}\times \hat{B}= mg \sin \Theta

or 

F= \vec{l}\times \vec{B}= (g \sin \Theta )\frac{m}{l}\rightarrow (1)

angle between rod and magnetic field is 90-\Theta 

i.e B \cos\Theta=mg \sin \theta

\Rightarrow ilB = mg \tan \theta  or i= \left ( \frac{m}{l} \right )\frac{g \tan \theta}{B}

i= \left ( \frac{0.5*10*}{0.25} \right )\frac{1}{\sqrt3}=20/\sqrt3= 11.3 amp

 

 

 

 


Option 1)

14.76 A

This is incorrect

Option 2)

5.98 A

This is incorrect

Option 3)

7.14 A

This is incorrect

Option 4)

11.32 A

This is correct

Posted by

Avinash

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