Question

If the vector field F=(x+2y+az)i+(2x-3y-z)j+(4x-y+2z)k is irrotational then the value of a is

Answer 1

Answer:

Step-by-step explanation: 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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