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Catalogs Discussion Forums -> Organic Chemistry -> dipole -> Go to message
This Post 10 points    (Olaaa!! Perrrfect answer.   in 2 votes )   [?]
2 replies   
If u are asking about electric dipole then answer is as below
 
Electric Dipole Moment
The electric dipole moment for a pair of opposite charges of magnitude q is defined as the magnitude of the charge times the distance between them and the defined direction is toward the positive charge. It is a useful concept in atoms and molecules where the effects of charge separation are measurable, but the distances between the charges are too small to be easily measurable. It is also a useful concept in dielectrics and other applications in solid and liquid materials.
Applications involve the electric field of a dipole and the energy of a dipole when placed in an electric field.
 
 
But for Magnetic moment it is as follows:
 
Magnetic moment can be explained by a bar magnet which has magnetic poles of equal magnitude but opposite polarity. Each pole is the source of magnetic force which weakens with distance. Since magnetic poles come in pairs, their forces interfere with each other because while one pole pulls, the other repels. This interference is greatest when the poles are close to each other i.e. when the bar magnet is short. The magnetic force produced by a bar magnet, at a given point in space, therefore depends on two factors: on both the strength p of its poles, and on the distance d separating them. The force is proportional to the product \mathbf{\mu}=\mathbf{p}\mathbf{d}, where \mathbf{\mu} describes the "magnetic moment" or "dipole moment" of the magnet along a distance R and its direction as the angle between R and the axis of the bar magnet.
Any rotating charged object, from quarks to galactic superclusters, has a magnetic moment.
For relation between magnetic moment and magnetization see magnetization.
Magnetism can be created by electric current in loops and coils so any current circulating in a planar loop produces a magnetic moment whose magnitude is equal to the product of the current and the area of the loop. When any charged particle is rotating, it behaves like a current loop with a magnetic moment.
The equation for magnetic moment in the current-carrying loop, carrying current \mathbf{I} and of area vector \vec{a} for which the magnitude is given by:
\vec{\mu}=\mathbf{I}\vec{a}
where
\vec{\mu}is the magnetic moment, a vector measured in ampere?square metres, or equivalently joules per tesla, \mathbf{I}is the current, a scalar measured in amperes, and \vec{a}is the loop area vector , having as x, y, and z coordinates the area in square metres of the projection of the loop into the yz-, zx-, and xy-plane
 
Catalogs Discussion Forums -> Algebra -> probability -> Go to message
This Post 5 points    (Olaaa!! Perrrfect answer.   in 1 votes )   [?]
2 replies   
Well the answer is (b)
 
The explanation is as follows:
 
Number of ways of selection of 8 letters from 24 letters of which 8 are a, 8 are b, is given by following cases
 
1) 8a = 1
2) (7a,1b) ; (7a,1 letter other than a and b) = 1 + 8C1 = 1+8 = 9
3) (6a,2b); (6a,1b,1 other than a and b); (6a, 2 other than a and b) = 1+8C1+8C2
                                                         = 9 + 28 = 37
4) (5a,3b) and so on = 1+8C1+8C2 + 8C3 = 37 + 56 =93
5) (4a,4b)...and so on = 1+8C1+8C2 + 8C3+ 8C4= 93+70 = 163
6) (3a,5b)...and so on = 1+8C1+8C2 + 8C3+ 8C41+8C5 = 163+56 = 219
7) (2a,6b)...and so on= 1+8C1+8C2 + 8C3+ 8C41+8C5 1+8C6=219+28 = 247
8) (a,7b)....so on = 1+8C1+8C2 + 8C3+ 8C41+8C5 1+8C6+8C7=247+8 = 255
 
Similarly we can interchange the role of 'a' and 'b' to get exactly eight more cases as above.
 
so total number of possible selections are = 2(1+9+37+93+163+219+247+255)
 
= 2 x 1024 = 211 = 8 x 28
Catalogs Discussion Forums -> Differential Calculus -> maxima and minima -> Go to message
This Post 5 points    (Olaaa!! Perrrfect answer.   in 1 votes )   [?]
3 replies   
Another approach could be as follows:

let ABC be a triangle inscribed in the circle with center O and radius r.

If the area of this triangle is maximum, then vertex C should be at a maximum distance from the base AB i.e. CD must be perpendicular to AB.

Hence, ABC is an isosceles triangle.
If  BCD = , where D is the mid-point of BC, then BOD = 2

so, AB = 2 BD

= 2r sin 2

CD = CO + OD = r + r cos 2

If S be the area of the triangle ABC, then

S = (1/2) AB x CD

= (1/2) x 2r sin 2 (r + r cos 2)

ds/d = r2[sin2 (-2 sin2) + (1 + cos2)(2 cos2)]

= 2r2[cos22 - sin22 + cos2] = 2r2(cos4 + cos2)

For maximum and mimimum

ds/d = 0

or, cos4 + cos2 = 0

or, 2 cos3 cos = 0

so, Either cos3 = 0 or, cos =0

If cos = 0 or 3 = /2 or = /6

(d2s/d2) = -ve, for = /6

ACB = 2 = 2(/6) = /3 = ABC = BAC

so ABC is an equilateral triangle.
Catalogs Discussion Forums -> Integral Calculus -> integrate -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]
4 replies   
I = (1-t)2dt

=
(1- 2t + t2) dt

= t - t2 + t3/3 + C

where C is constant of integration.
Catalogs Discussion Forums -> Mechanics -> Kinematics -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]
4 replies   
Displacement and distance of a particle will be same provided the particle is having a rectilinear motion or it is moving in a straight line.

Thus the displacement  and distance will be same only assuming that particle is moving in a straingt line, and is obtained by putting t = 5 in the expression of S.

Acceleration ofcourse is not constant in this case.

Catalogs Discussion Forums -> Mechanics -> GRAVITATION -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]
2 replies   
Actually center of mass is not the point through which we hold the body. Thus, your perception for center of mass needs to be changed.

It is infact the point at which the whole mass of body is assumed to be concentrated or it may also be referred as center of gravity.
Catalogs Discussion Forums -> Electricity -> physic - electro static flux -> Go to message
This Post 2 points    (Olaaa!! Perrrfect answer.   in 1 votes )   [?]
3 replies   
First of charge doesn't have units of micro farad it must be micro coulomb.
 
Now consider cube of side 10cm enclosing the charge of 10 micro coulomb at the centre which is at a distance 5cm from each of the sides.
 
Now applying gauss's law of electrostatics
 
Electric Flux passing through the cube = q/0.
 
But by symmetry of the problem the flux is equidistributed through each sides.
 
Thus flux through the square plate of side 10cm = q/60
 
or flux = 10^(-6)/60.
 
Catalogs Discussion Forums -> Algebra -> complex numbers -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]
7 replies   
Dear sumeet i think your approach is fair enough and ok.
Catalogs Discussion Forums -> Integral Calculus -> DEFINITE INTEGRAL -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]
3 replies   
Is your Integral this?

e(t^2-2)/2 .[(1/t) + t log t)]dt
Catalogs Discussion Forums -> Integral Calculus -> indefinite integral -> Go to message
This Post 5 points    (Olaaa!! Perrrfect answer.   in 1 votes )   [?]
2 replies   
 I = (x2.cosx)/(1+sinx)2 .dx   

Integration by parts gives

I =
x2cosx/(1+sinx)2 .dx  - (2x.cosx)/(1+sinx)2 .dx

=
x2cosx/(1+sinx)2 .dx - 2xcosx/(1+sinx)2 .dx  + 2cosx/(1+sinx)2 .dx

put  (1+sinx) = t, then cosx dx = dt

Now you can proceed.
Catalogs Discussion Forums -> Algebra -> COMPLEX NOS -> Go to message
This Post 5 points    (Olaaa!! Perrrfect answer.   in 1 votes )   [?]
3 replies   
Well done krish. The approach and solution suggested by krish is perfect. And

the answer is undoubtedly w2

Now lets take the QUERY of ariyam66

If we take x = y = z = k (say)

then the expression becomes

(xA + yB + zC)/(xB + yC + zA) = (x + yw + zw2)/(xw + yw2 + z)

= k(
1 + w + w2) / k (w + w2 + 1)

=
(1 + w + w2) /  (w + w2 + 1)

= 0/0 and this can not be equated to 1

As 0/0 is not defined.
Catalogs Discussion Forums -> Modern Physics -> Differential calculus -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]
6 replies   
dy/dx+ y tan x = xm  cos x

To solve the equation i suggest the following approach

this is first order linear differential equation

and Integrating factor = IF = tanx dx

or IF = log Isec xI

Thus solution of such equations is given by

y(IF) =
xm  cos x (IF) dx

or y = [1/
log Isec xI ] xm  cos x log Isec xI dx

Now try to solve the above integral by reduction formula.
Catalogs Discussion Forums -> Magnetism -> NPN transistor -> Go to message
This Post 12 points    (Olaaa!! Perrrfect answer.   in 3 votes )   [?]
2 replies   
I expect your answer on npn transistor to be more specific.
 
I think u want to enquire about
 
 How does a transistor work?
 
Answer) The design of a transistor allows it to function as an amplifier or a switch. This is accomplished by using a small amount of electricity to control a gate on a much larger supply of electricity, much like turning a valve to control a supply of water.

Transistor terminalsTransistors are composed of three parts ? a base, a collector, and an emitter. The base is the gate controller device for the larger electrical supply. The collector is the larger electrical supply, and the emitter is the outlet for that supply. By sending varying levels of current from the base, the amount of current flowing through the gate from the collector may be regulated. In this way, a very small amount of current may be used to control a large amount of current, as in an amplifier. The same process is used to create the binary code for the digital processors but in this case a voltage threshold of five volts is needed to open the collector gate. In this way, the transistor is being used as a switch with a binary function: five volts - ON, less than five volts - OFF.
Catalogs Discussion Forums -> Magnetism -> biot savarts law -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]
3 replies   
the derivation part is avilable in any standrd txt or ref. book for class 12. so above i have presented the law only
Catalogs Discussion Forums -> Magnetism -> biot savarts law -> Go to message
This Post 2 points    (Olaaa!! Perrrfect answer.   in 1 votes )   [?]
3 replies   
The Biot-Savart law is a physical law with applications in electromagnetics and, more generally, it is also useful in numerous other applications of vector field analysis. As originally formulated, the law describes the magnetic field set up by a steady current density.
 Mathematically, the Biot-Savart law provides an inverse to the curl operation; the result is unique up to gauge transformation. As such it has numerous additional applications.
The Biot-Savart law is fundamental to magnetostatics just as Coulomb's law is to electrostatics. The Biot-Savart law follows from the Lorentz transformations of the electric field of a point-like electric charge, which results in a magnetic field, and is fully consistent with Ampère's law, much as Coulomb's law is consistent with Gauss' law.
In particular, if we define a differential element of current
I dmathbf{l}
then the corresponding differential element of magnetic field is
dmathbf{B} = K_m rac{I dmathbf{l} 	imes mathbf{hat r}}{r^2}
where
K_m = rac{mu_0}{4pi} ,, where ?0 is the magnetic constant
Imathbf{} is the current, measured in amperes
dmathbf{l} is the differential length vector of the current element
mathbf{hat r} is the unit displacement vector from the current element to the field point and
rmathbf{} is the distance from the current element to the field point
 
 
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