CBSE-XI-Physics
35: Magnetic Field due to a Current
- #2A current-carrying, straight wire is kept along the axis of a circular loop carrying a current. This straight wire
(a) will exert an inward force on the circular loop
(b) will exert an outward force on the circular loop
(c) will not exert any force on the circular loop
(d) will exert a force on the circular loop parallel to itself.digAnsr: cAns : (c) will not exert any force on the circular loop
The magnetic force on a wire carrying an electric current i is given as​ `` \stackrel{\to }{F}=i.(\stackrel{\to }{l}\times \stackrel{\to }{B})``, where l is the length of the wire and B is the magnetic field acting on it. If a current-carrying straight wire is kept along the axis of a circular loop carrying current, then according to the right-hand thumb rule, the magnetic field due to the wire on the current-carrying loop will be along its circumference, which contains a current element `` id\stackrel{\to }{l}``.
So, the cross product will be​
`` (\stackrel{\to }{l}\times \stackrel{\to }{B})=0``
`` \Rightarrow \stackrel{\to }{F}=0``
Thus, the straight wire will not exert any force on the loop.
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