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  • How to solve this problem- A solid metal cube of edge length 2cm is moving in a positive y-direction at a constant speed of 6m/s. There is a uniform magnetic field of 0.1 T in the positive z-direction. The two faces of the cube perpendicular to the x-axis

A solid metal cube of edge length 2cm is moving in a positive y-direction at a constant speed of 6m/s. There is a uniform magnetic field of 0.1 T in the positive z-direction. The two faces of the cube perpendicular to the x-axis, is:

  • Option 1)

    12mV

  • Option 2)

    6mV

  • Option 3)

    1 mV

  • Option 4)

    2mV

Answers (1)

best_answer

 

Force on a charged particle in magnetic field -

\underset{F}{\rightarrow}=q (\underset{V}{\rightarrow}\times \underset{B}{\rightarrow})

F=qVB\sin \theta

- wherein

\underset{V}{\rightarrow} - velocity of the particle 

\underset{B}{\rightarrow}  magnetic field 

 

Given

 

l=2cm

V=6m/s

 

\varepsilon _{x}=Emf =P.D. \: between \: two\: faces\: of\: cubes\: perpendicular\: to\: x-axis\\\\so\: \: \varepsilon _{x}=l\left ( \vec{V}\times \vec{B} \right )\\\\=\frac{2}{10^{2}}\left ( \left ( 6\times 0.1 \right ) \right )\\\\\varepsilon _{x}= \Delta V=\frac{12}{103}V\\\\\varepsilon _{x}=12mV

 


Option 1)

12mV

Option 2)

6mV

Option 3)

1 mV

Option 4)

2mV

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