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Q.

A wire ABCDEF (with each side of length L) bent as shown in the figure and carrying a current I is placed in a uniform magnetic induction B parallel to the positive y-direction.  Find the force experienced by the wire and direction?


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a

ILB, +ve z-axis

b

ILB, -ve z-axis

c

-ILB, +ve z-axis

d

Zero  

answer is A.

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Detailed Solution

This force will act along z axis so, ILB is Z axis.
The expression of force is F= I (l × B)
The angle between  FE and B is 1800
BA and B is 0
For both of these L × B=0
The angle between ED and B is 900
Force on ED is L ×  B=ILB sin 900
                                      =ILB
The angle between CB and B is 2700
Force on CBis L ×  B=ILB sin 2700
                                    =-ILB
The force on ED or CB acts perpendicular to both ED and B or CB or Btowards x-axis. These two factors cancel each other as they act opposite
The angle between DC and B is 900
Force on  DC is L × B=ILB sin 900 
                                      =ILB
this force will act along z axis so, ILB is Z axis.
 
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