**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| =(x^{2}+y^{2}+z^{2})

Then gard |r| ∆| r| (x^{2} +y^{2} +z^{2})_{1/2} =r

Where ∆ =

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

curl of =∆ .r =0

**Problem 2.** If ϕ = 5xy- 3y^{2} 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| =√ (x^{2}+y^{2}+z^{2})

Now ∆ ^{2} (1/r ) = ∆ .{ ∆ ( 1/r) }

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

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

Hint: W= ʃ f .dr = ʃ f_{1} dx +f_{2} dy

F_{1} =(2xy) ,f_{2} =y_{3}

W =ʃ (2xy dx ++y_{3} dy)

Y =x _{2} : dy =2x dx

W = ʃ ( 2x x^{2} dx –x^{6} .2x dx) = ʃ^{2}_{0} (2x^{3} -2x^{7} ) 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 ^{n}_{m} (f _{x} d _{y} +f _{y} d y +f _{z} dz) ; x-y plane mean z =0

ʃ^{2}_{0} 6x dx +ʃ ^{2}_{0} 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 ʃ_{s} curl F. dS = 0 for any closed surface.

**Hint**: Gauss’s theorem ʃ ʃ_{s} 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 + z^{3}) i + x^{2} 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 Nm^{2} L^{-1}.

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

E_{x} = (ax + by + c), E_{Y} = 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 =al^{3}

From the Gauss’s law in electrostatics

ʃ ʃ _{s} E .ds =q /ε_{0}

from equations (i) and (ii) ,a L_{3} =q / ε_{0} àq =ε_{0} a L_{3}

**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/m^{2} which is perpendicular to the plane of the ring. If the magnetic field is decreasing at the rate of 0.1 Weber/m^{2} 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 =πr^{2} =3.14 x (10 x 10^{-2})^{2} =3.14 x 10^{-2} m^{2} , Db/dt =0.1/1 weber /m^{2} 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/ε_{0}, where A = area of each plate of the capacitor àq=ε_{0} EA

àdq/dt =ε_{0} A De/dt ài_{d } ε_{0 } A DT/dt where i_{d } = displacement current

We know that q =C V à dq/dt =C dv/dt à i_{d } = C Dv /dt

Alternatively: i_{d} =ε_{0} A de/dt à i_{d} = ε_{0} A/d/dt (Ed) ; d =spacing of plates

i_{d} =ε_{0} A/d .dv/dt (;V =E.D)

i_{d} = C Dv/ dt (;C =ε_{0} 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 /2ε_{0}r = 2λ/4πε_{0}r =9 x 10^{9} 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 r^{2} = 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 =√ µ_{0}ε_{)} =376.2

Eq .(i) x Eq .(ii) à E^{2} =1000/16π x 376.2 à E =48.87 V/m