a current carrying closed loop in the form of a right angle isosceles triangle ABC is placed in a uniform magnetic field acting along AB. If the magnetic forces on the arm BC is F ;the force on the arm AC is
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If the magnetic field along AB is B , the force acting on BC=F=BiL
Now the length of the side AC is Root"2L"
Also the current reverses the direction in this section as compared to that in BC.
Hence force on AC= Root"2"L( B× I )
= − Root"2"BILsin(45)
=−BIl=− F
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