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Question

A long metal bar of 30 cm length is aligned along a north south line and moves eastward at a speed of 10 ms1. A uniform magnetic field of 4.0 T, points vertically downwards. If the south end of the bar has a potential of 0 V, the induced potential at the north end of the bar is
  1. +12 V
  2. 0 V
  3. 12 V
  4. cannot be determined since there is not closed circuit


Solution

The correct option is A +12 V
The motional emf produced due to velocity v of rod is given by
ε=vBl ...(i)
where l=length of rod perpendicular to its velocity
since length of rod is making angle 90 with east (Direction of velocity of rod), hence
l=l=30 cm=0.30 m ...(ii)
Substituting values:
ε=vBl
ε=10×4×0.30=12 V
The sign of Motional emf (ε) is given by:
v×B, so using right hand rule for cross product,the direction of thumb will point towards +ve terminal.
direction of v is ^i, direction of B is ^k
^i×^k=^j i.e along +ve  y direction (NORTH)  for (+ve) terminal of motional emf.
ε=VNVS
12 V=VN0
VN=+12 V

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