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Numerical Problems Based on Electromagnetism

Problem 1. Find the gradient, divergence and curl of | r |, where r is the position vector

 

Hint :              r=I x +l y + k z | r| =(x2+y2+z2)

 

Then  gard  |r|              ∆| r| (x2 +y2 +z2)1/2   =r

Where ∆ =

divergence of r = ∆.r = . (I x +j y+ k z) =3

curl of =∆ .r =0

 

Problem 2. If ϕ = 5xy- 3y2 z:3, find ‘V$ at the point (-1, 2, -1).

Hint : ∆ ϕ =

 

è At  (-1,2 ,-1) ∆ ϕ = 10i +7j – 36k.

 

Problem 3 . If  A =3xyz2 i+2xy3 j –x2 yz k , find ∆ .A at a point (1,-1,1)

 

HINT : ∆ A =  .(3xyz2 I +2xy3 j –x2yz k) at ( 1,-1,1)

 

Problem 4. Prove that  ∆2 (1/r) =0

 

Hint : here                   r =ix +jy +kz ,then r  =|r| =√ (x2+y2+z2)

Now                            ∆ 2 (1/r ) = ∆ .{ ∆ ( 1/r) }

 

Problem 5. Find the work done in moving a particle along the curve

Y =x2 by a force f  =I (2xy ) –j (y3) from (0,0) to (2,4)

Hint:                W= ʃ f .dr = ʃ f1 dx +f2 dy

F1 =(2xy) ,f2 =y3

W =ʃ (2xy dx ++y3 dy)

Y =x 2 : dy =2x dx

W = ʃ ( 2x x2 dx –x6 .2x dx) = ʃ20 (2x3 -2x7 ) dx = -56 unit

 

Problem 6. Evaluate the line integral of function F = i (6x) + j ( 4y) between the points  (0, 0) and (2, 2) in x-y plane

Hint : ʃ f .dl = ʃ nm (f x d y +f y d y +f z dz) ; x-y plane mean z =0

ʃ20 6x dx +ʃ 20 4y dy =20

 

Problem 7. Obtain the values of the following:

 

(i) curl grad ϕ              (ii) grad div A and

(iii) div curl A.

 

Hint: r) 0

 

(iii) 0.

 

Problem 8. Prove that ʃ ʃs curl F. dS = 0 for any closed surface.

 

Hint: Gauss’s theorem  ʃ ʃs curl F. dS = ʃ ʃ ʃ V.(curl F) dV v

= ʃ ʃ ʃ  div.curl F dV

v

div curl F = 0

 

Problem 9. Using Stoke’s theorem, prove that

= 0, where r is position vector.

 

Hint:  = ʃ ʃ  curl A.dS, here curl r = 0

 

Hence                          ʃ  r.dl = 0.

 

Problem 10. Electric field associated with a charge body is given by E = 4i + j + 7k  and the surface is given by S = 8i + 3j. Calculate the electric flux coming out the surface.

 

Hint: ϕ = E.S = (4i + j + 7k).(8i + 3j) = 35 unit.

 

Problem 11. Show that F = (2xy + z3) i + x2 j + 3xz k is a conservative field.

 

Hint:    ∆’ x F = 0 , Hence F is’ conservative.

 

Problem 12. Examine whether the field given byE= a (xy2 i + y3 j) is conservative. (here i and j are unit vectors), a is constant

 

Hint :   ∆ x E = – 2 ax y k ; ∆ x E  =0

Hence field is not conservative.

 

Problem 13. If E = 3i + 4j + 8k. Calculate the electric flux through a surface area 100 units lying in x-y plane.

 

Hint:                ϕ = E.S = (3i + 4j + 8k). 100 k

= 800 Nm2 L-1.

 

Problem 14. The electric field E in a certain space is given by

Ex = (ax + by + c), EY = 0, E z = 0.

 

Use Gauss’s theorem, to evaluate the charge enclosed in a cube with one of its sides of  length L along the x-axis, the two faces of the cube are perpendicular to the x-axis lying in the planes x = 1 and X=l+L.

 

Hint: According to Gauss theorem of divergence

 

ʃ  ʃ E .ds = ʃ ʃ ʃ div E dv

 

= ʃ ʃ ʃ a dv =a v =al3

 

From the Gauss’s law in electrostatics

ʃ ʃ s E .ds =q /ε0

from equations (i) and (ii) ,a L3 =q / ε0  àq =ε0 a  L3

 

Problem 15. A long wire carries a current of 5 mA. Find the integral of B around the path of radius 10 cm, enclosing the wire. µ0/4π =10 -7 S. I units.

 

Hint :

 

Problem 16. A ring of radius 10 cm and resistance 2 Ohm is placed in a magnetic field  of 0.5 Weber/m2 which is perpendicular to the plane of the ring. If the magnetic field is decreasing at the rate of 0.1 Weber/m2 per second, find the values of the induced cmf and the induced current in the ring.

Hint: The time rate of change of magnetic flux is

Dϕ /dt = d / dt (BA) =A Db/dt

A  =πr2 =3.14 x (10 x  10-2)2 =3.14  x 10-2 m2 , Db/dt =0.1/1 weber /m2 sec

Dϕ /dt =3.14 x 10-3 weber / sec

 

According Faraday’s law e = dϕ/dt =3.14 x 10-3 volts

 

If R is resistance of the loop, then induced current

I= e/R =3,14 /2  =1.57  Ma

 

Problem 17. Express displacement current, appeared inside a parallel plate capacitor, in terms of electric field (E) and electric potential.

Hint: Frame Gauss’s law

 

Assuming the Gaussian surface enclosing the charges of in the plate, then above

 

equation becomes EA = q/ε0, where A = area of each plate of the capacitor àq=ε0  EA

 

àdq/dt =ε0 A De/dt àid  ε0  A DT/dt where i = displacement current

 

We know that q =C V  à dq/dt =C dv/dt à id  = C Dv /dt

Alternatively:  id0 A de/dt à id = ε0 A/d/dt  (Ed) ; d =spacing of plates

id0 A/d .dv/dt                      (;V =E.D)

id = C Dv/ dt                           (;C =ε0 A/d)

 

Problem 18. An infinite long wire is stretched horizontally 4 m above the surface of  the It has a charge of earth. 1 ).1 C/cm. Calculate electric field at a point on earth vertically below the wire.

Hint : E = λ /2ε0r = 2λ/4πε0r =9 x 109 x 2 x 10 -4 /4 = 4.5 x 105 N /Coul .

 

Problem 19. Assuming that all the energy from a 1000 W lamp is radiated uniformly, calculate the average values of the intensities of electric and magneticf ields of radiation at a distance of 2 m from the lamp.

 

Hint: Energy of lamp = 1000 W; Area illuminated = 4 π r2 = 4 π (2)2 = 16n met2

 

Energy radiated per unit area per second =1000 /16π

Hence from Poynting theorem  |S| = |E x H | = EH =1000/16π

And                 E/H =√ µ0ε) =376.2

Eq .(i) x Eq .(ii) à  E2 =1000/16π x 376.2 à E =48.87 V/m