I vector field F- (2y + 2) ++ (2x + 2)]+yk is irrotational then value of a is
O
A 0
B
1
5
D
4

December 30, 2021

By Jasmine

1 thought on “I vector field F- (2y + 2) ++ (2x + 2)]+yk is irrotational then value of a is<br />O<br />A 0<br />B<br />1<br />5<br />D<br />4<b”

1. Answer:

   F¯¯¯¯$=$(x+2y+az)i+(bx−3y−z)j+(4x+cy+2z)k

   and

   r¯¯=xi¯+yj¯+zk¯¯¯

   ∴dr=dxi¯+dyj¯+dzk¯¯¯

   Since F¯¯¯¯ is irrotational

   Curl F¯¯¯¯=0

   ∴ ∣∣∣∣∣i¯∂∂xFxj¯∂∂yFyk¯¯¯∂∂zFz∣∣∣∣∣=0

   ∴ ∣∣∣∣∣i¯∂∂xx+2y+azj¯∂∂ybx−3y−zk¯¯¯∂∂z4x+cy+2z∣∣∣∣∣=0

   ∴i[∂∂y(4x+cy+2z)− ∂∂z(bx−3y−z)]−j[∂∂x(4x+cy+2z)+ ∂∂z(x+2y+az)]+k[∂∂x(bx−3y−z)− ∂∂y(x+2y+az)]

   ∴i[c−1]−j[4−a]+k[b−2]=0i+0j+0k

   Comparing co-efficient of i,j,k

   ∴[c−1]=0

   ∴[4−a]=0

   ∴[b−2]=0

   ∴c=1 , a=4 , b=2

   ∴ F¯¯¯¯=(x+2y+az)i+(bx−3y−z)j+(4x+cy+2z)k

   ∵F¯¯¯¯ is irrotational ,there exists a scalar potential of F¯¯¯¯ such that F¯¯¯¯= ∇∮

   ∴(x+2y+az)i+(bx−3y−z)j+(4x+cy+2z)k=i∂∮∂x+j∂∮∂y+k∂∮∂z

   Comparing co-eeficient of i,j,k we get,

   ∂∫∂x= (x+2y+az)

   ∂∫∂y= (bx−3y−z)

   ∂∫∂z= (4x+cy+2z)

   Integrating we get ,

   ∮=(x22+2yx+4zx)+(2xy−3y22−zy)+(4zx−2y+2z22) + c

   As the same term is twice, it is written only once in this type of integration.

   ∴ ∫=x22+2yx+4zx+2xy−3y22−zy+4zx−2y+z2+c

   ∴ Scalar potential of F¯¯¯¯= ∫= 1/2[x2+4yx+8zx−3y2−2zy+2z2+c

   Now work done in moving a particle in this field = ∫CF¯¯¯¯.dr¯¯¯¯¯

   =∫C((x+2y+4z) dx+(2x−3y−z)dy+(4x+y+2z)dz

   Integrating we get ,

   =[x2+4yx+8zx−3y2−2zy+2z2](3,3,2)(1,2,−4)

   =[9/2+2∗3∗3+4∗2∗3−3∗9/2−3∗2∗−4]−−[1/2+2∗2∗1+4∗(−4)∗1−3∗2−2∗(−4)+16]=24.5

   ∴Work done in moving a particle=24.5

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