All quantities that have magnitude and direction are known as vectors. A vector will be represented by a bold face letter (i.e., with thick capital letter). The force F which is a vector will be represented by F. The alternate way of representing the force vector is by putting an arrow as F .

**1.6.1. Magnitude of a Vector**. The magnitude of a quantity is always a positive number. Thus the magnitude of a quantity(- 40) is+ 40 units. The magnitude of a quantity mathematically is represented by a set of vertical lines enclosing the quantity. Hence mathematically the magnitude of a quantity(- 40) is represented as :

| – 40 units | = + 40 units

Similarly, the magnitude of a vector is a positive number of units corresponding to the length of the vector in those units. Hence magnitude of a vector A is represented as :

Magnitude of vector A = | A|

where | A | is a positive scalar quantity. The magnitude* is also represented by A (i.e., capital letter A).

Magnitude of a vector A = I A I =A.

**1.6.2. Multiplication of a Vector by a Scalar**

Let A = a given vector

m = a given scalar

Vector A multiplied by scalar m = mA

where mA = a vector having same direction as A and a magnitude equal to the ordinary scalar product between magnitudes of m and A.

**1.6.3. Vectorial Representation of Forces and Moments**

**(A) Vectorial representation of a force**. Fig. 1.15 shows a force F acting at the origin 0. Let the magnitude of Y this force is equal to length OA. Then this force vector F is represented by vector OA. Through point A, draw planes parallel to co-ordinate planes.

These planes along with co-ordinate planes make a rectangular box. The force F is then represented by the diagonal of the box and its three components Fx, FY and Fz by its edges.

Let θ _{x} = Angle made by force F with x-axis

θ= Angle made by force F withy-axis

θ _{z} = Angle made by force F with z-axis

Now f _{x} = f cos θ _{x}

F _{y} = f cos θ _{y}

F _{z }= f cos θ _{z}

The cosines of θ x, θ y and θ z are known as the direction cosines of the force F and denoted by

l = cos θ x , m = cos θ y and n = cos θ z

The three angles* are related by

Cos^{2} θ x + cos ^{2} θ y + cos^{2} θ z = 1

l^{2} + m^{2} + n^{2 }= 1

Let i = vector of unit length in the positive x-direction,

j = vector of unit length in the positive y-direction,

and k = vector of unit length in the positive z-direction.

Then force vector F is represented by

F = f x I + f y j + f x k

Magnitude of the vector F is given by

But F _{x} = F cos ex, F_{Y} = F cos θ y and F z = F cos θz. Substituting these values in equation (1.9), we get

F = (F cos 9) i + (F cos 0) j + (F cos 9z) k … (1.10)

**Unit Vector**

The vector having a unit length (or unit magnitude) is known as unit vector. Hence unit vector corresponding to force vector F is equal to F +Magnitude of vector F.

Unit vector = force vector f / magnitude of vector f = f / f

**Note.** Force vector is represented by THICK, BOLD FACE and magnitude is represented by capital letters only.

**Response time : t _{0}** : ‘ This is the time during which <artier concentration reduces to of starting carrier concentration.

Hence at t = t_{0} , n = n_{0} /2

Using equation (5) 1 / n_{0}/2 = A t_{0} + 1 /n_{0}

Response time is directly proportional to σat a given light intensity level. Good photoconductors have large t0. These are rarely found in practice.

**(b) Sensitivity or Gain Factor (G):** This can be calculated as the ratio of numbers of carriers crossing the unit area of specimen to the number of photons absorbed by that area in the specimen. It can be represented by ‘ G’. So for a specimen of thickness, ‘d’ and the cross section area unity, we have

G = Particle flux /L x (d x1)

We know flux is the rate of flow of carriers per unit area per unit time is given by ‘Fn

Also the current density

J _{n} = σ E = n_{0} e µ (V /d)

A potential V produces the particle flux F _{n} is given by

F _{n} = µV /d n0

F _{n} = µV /d (L t0)

F _{n} = µV /d x L x

Hence F _{n} = µV /d x L x

So from equation (8), G =F _{n} / L d

G = 1 /L d x µV /d x L x

G = µV / d^{2} (L A)1/2

Now, let T _{e} is the life time of the electron before illumination and it is nothing but the response time t_{0} , so

Let T_{d }is the transit time of an electron between the electrodes and is given by

Td = t_{0} = d / µE

So from equation (10)

The expression of G as given in equation (11) is quite general, however, this theoretical value of t0 do not agree and in some instances the discrepancy is = 108 times. The result then indicates.

• Model is failure

• We should look out for some missing phenomenon not considered in this expression.