1000 Solved Problems in Modern Physics

1.3 Solutions 35

(b) If the field is solenoidal, then,∇.rF(r)= 0

∂(xF(r))

∂x

+

∂(yF(r))

∂y

+

∂(zF(r))

∂z

= 0

F+x

∂F

∂x

+F+y

∂F

∂y

+F+z

∂F

∂z

= 0

3 F(r)+x

∂F

∂r

x

r

+y

∂F

∂r

y

r

+z

∂F

∂r

z

r

= 0

3 F(r)+

(

∂F

∂r

)(

x^2 +y^2 +z^2

r

)

= 0

But (x^2 +y^2 +z^2 )=r^2 , therefore,∂∂Fr=−^3 Fr(r)

Integrating, lnF=−3lnr+lnCwhereC=constant

lnF=−lnr^3 +lnC=ln

C

r^3

ThereforeF=C/r^3. Thus, the field isA=

r

r^3

(inverse square law)

1.6 x=t,y=t^2 ,z=t^3

Therefore,∫ y=x^2 ,z=x^3 ,dy= 2 xdx,dz= 3 x^2 dx

c

A.dr=

∫

(yˆi+xzˆj+xyzkˆ).(ˆidx+ˆjdy+kˆdz)

=

∫ 1

0

x^2 dx+ 2

∫ 1

0

x^5 dx+ 3

∫ 1

0

x^8 dx

=

1

3

+

1

3

+

1

3

= 1

1.7 The two curvesy =x^2 andy^2 = 8 xintersect at (0, 0) and (2, 4). Let us

traverse the closed curve in the clockwise direction, Fig. 1.6.

∫

c

A.dr=

∫

c

[(x+y)ˆi+(x−y)ˆj].(iˆdx+ˆjdy)

=

∫

c

[(x+y)dx+(x−y)dy]

=

∫ 0

2

[(x+x^2 )dx+(x−x^2 )2xdx] (alongy=x^2 )

Fig. 1.6Line integral for a
closed curve