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An insulating thin rod of length l has a linear charge density p(x)=P_{0}\frac{x}{l} on it. The rod is rotated about an axis passing through the origin (x=0) and perpendicular to the rod. If the rod makes n rotations per second, then the time-averaged magnetic moment of the rod is 

  • Option 1)

      \pi n p l^{3}

      

  • Option 2)

    \frac{\pi }{3}npl^{3}

  • Option 3)

    \frac{\pi }{4}npl^{3}

  • Option 4)

    npl^{3}

Answers (1)

best_answer

 

Magnetic moment (M) -

M=NiA

- wherein

N-number of turns in the coil 

i-current throughout the coil 

A-area of the coil 

Assuming a dx element at a distance x from point O.

Then dq=\lambda dx

A=\pi x^{2}

M=NIA

dm = \int_{0}^{l}(\rho _{0}\frac{x}{l}).n.dx.\pi x^{2}

M=\frac{n\rho _{o}\pi }{l}\int_{o}^{l}x^{3}=\frac{n\rho _{o}\pi }{l}\frac{l^{4}}{4}

M =\frac{\pi }{4}n\rho _{o}l^{3}

 

 

 


Option 1)

  \pi n p l^{3}

  

Option 2)

\frac{\pi }{3}npl^{3}

Option 3)

\frac{\pi }{4}npl^{3}

Option 4)

npl^{3}

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