  Question

# A long metal bar of 30 cm length is aligned along a north south line and moves eastward at a speed of 10 ms−1. 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+12 V0 V−12 Vcannot be determined since there is not closed circuit

Solution

## The correct option is A +12 VThe 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=VN−VS 12 V=VN−0 ∴VN=+12 V  Suggest corrections   