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51.

Consider a conductor of length 40 cm where a potential difference of 10V is maintained between the ends of the conductor. Find the mobility of the electrons provided the drift velocity of the electrons is `5xx10^(-6) ms^(-1)`

Answer» Given `L = 40 cm, V = 10V, nu_d = 5xx10^(-6) ms^(-1).`
To find the electron mobility, we need the value of the
electric field, which can be obtained using the following
formula:
`E = (V)/(I) or E = (10)/(0.4) = 25Vm^(-1)`
Also the formula used for electron mobility is
`mu = (nu_d)/(E) = (5xx10^(-6))/(25) = (1)/(5)xx10^(-6)`
`=0.2xx10^(-6) = 2xx10^(-7) m^2 V^(-1) s^(-1)`
Discussion
52.

A potential difference V is applied to copper wire fo diameter d and lengt L, what will be the effect on, the electron drift speed by doubling (a) voltage V, (b) length L, and (c ) diameter d?

Answer» According to elerctron theory of metals, the dirft speed of an
electron inside a metal in the presence of an electric field E is
given by `v_d = ((etau)/(m)) E = ((etau)/(m))((V)/(L)) [ as E= (V)/(L)]`
a. As `vd propV,` on doubing V, drift velcoity will be doubled.
b.As `v_d prop(1//L),` on doubling L, drift velcoity will be halved.
c. As dirft veocity is independent of diameter d, it will not
charge on doubling the diameter.
Discussion
53.

Shown a conductor of length `l` having a circular cross section. The radius of cross section varies linearly form `a to b`. The resistivity of the material is`(rho)`.Assuming that `b-altltl`,find the resistance of the conductor.

Answer» Let at distacne x, the radius is r. Then,
`r = a +Deltar = a+((b-a)/(l))x`
`R= intdR = int(rhodx)/(pir^2) = int_0^l (rhodx)/(pi[a+((b-a)/(l))x]^2)`
`.`
Let `a + ((b-a)/(l)) x = z or ((b-a)/(l))dx = dz`
when z goes from a to b, we get
`R= intdR = int_a^b (rhol)/(pi(b-a)) (dz)/(z^2) = (rhol)/(pi(b- a)) int_a^bh (dz)/(z^2)`
`=(rhol)/(pi(b-a)) [(1)/(a)-(1)/(b)]=(rhol)/(piab)`
Discussion
54.

A copper coil has resistance of `20.0 Omega at 0^@C` and a resistance of `26.4 Omega at 80^@C`. Find the temperature coefficient of resistance of copper.

Answer» `R_(80^@C) = R_(0^@C)[1 + alpha Delta T]`
or `26.4 = 20.0 [1+alphaxx(80 - 0)] or (26.4)/(20) = 1+80 alpha`
On solving, we get ` alpha = 4xx10^(-3^@)C^(-1).`
Discussion
55.

A copper wire fo cross-sectional area `3.00xx10^(-6) m^2` carries a current 10.0A. a. Find the drift speed of the electrons in the wire. Assume that each copper atom contributes one free electron to the body of material. b. Find the average time between collisions for electrons in the copper at `20^@C.` The density of copper si `8.95 gcm^(-1)`, molar mass of copper is `63.5 gmol^(-1)`, Avogadro number is `6.02x10^(23)` electrons per mol and resistivity of copper is 1.7 xx10^(-8) Omegam.`

Answer» The volume occupied by 63.5 g of copper is
`V = (M)/(rho) = (63.5)/(8.95) = 7.09 cm^3 mol^(-1)`
As each copper atom contributes one free electron to the body of the meterial, the density of freee electrons is
`n = (6.02xx10^(23))/(7.09xx10^(-6)) = 8.48xx10^(28)` electrons per cubic meter
The drift speed is
`v_d = (I)/("ne"A)`
`=(10.0)/(8.48xx10^(28)xx1.60xx10^(-19)xx3xx10^(-6)) = 2.46xx10^(-4)ms^(-1)`
b. Average time between collision of electrons is
`tau = (m_e)/("ne"^2rho) = (9.10xx10^(31))/(8.48xx10^(28)xx(1.6xx10^(19))^(2)xx1.7xx10^(-8))`
`=2.5xx10^(-14)S`
Discussion
56.

A straight conductor of uniform cross section carries a time varying current, which varies at the rate dI//dt = I. If s is the specific charge that is carried by each charge carries of the conductor and l is the length of the condcutor, then the totalt force experienced by all the charge carries per unit length of the conductor due to their drift velocities only isA. `F = I s`B. `F = (I)/(2sqrtls)`C. `F = (I)/(s)`D. `F = (2I l)/(s)`

Answer» Correct Answer - C
Momentum of each charge carrier moving with a drift velocity
v is mv. Total number of charge carries in the sample is
N = n (Al), where n is the number of charge carries per unit
volume and A is area of cross section of the conductor.
Total momentum is p = N(mv)=nAlmv
Further, we have
`upilon = I//"ne"A`
or `P = nAim((I)/(meA)) = l((M)/(e))I`
Since `F = (dp)/(dt)`
or `F=(l)/(S) I or (F)/(l) = (I)/(s)` [ `:.` specific charge = e/m]
Discussion
57.

A copper wire having cross- sectional area of `0.5 mm^2` and a length of ` 0.1 m` is initially at `25^@C` and is thermally insulated form the surrounding. If a current of `10A` is set up in this wire, (a) find the time in which the wire will start melting. (b) What will be the change of resistance with time, if the length of the wire is doubled? For copper, resistivity is `1.6xx10^(-8)Omega m,` density is `9,000 kgm^(-3)` and specific heat capacity is `0.09 cal g^(-1)C^(-1).` Melting temperature of copper is `1075^@C`

Answer» `a=0.5mm^(2)=0.5xx10^(-6)m^(2),l=0.1m,T_(1)=25^(@)C,I=10A`
`T_(2)=1075^(@)C`
a. `I^(2)Rt=mstriangleT` or `I^(2)rho(l)/(a)xxt=mstriangleT`
or `t=(mstriangleTxxa)/(I^(2)rhol)=((d xx axxI)striangleTxxa)/(I^(2)rhol)=(da^(2)triangleT)/(rhoI^(2))`
`9xx10^(3)xx(0.5xx10^(-6))^(2)`
`=(9xx10^(-2)xx10^(3)xx4.18xx1050)/(10xx10xx1.6xx10^(-8))`
`=555.5s=9min15s`
(b). Since length does not occur in the expression of time. the melting does ot depend on the length. So time taken will be the same.
Discussion
58.

How much time will be taken by an electron to move distance I = 1km in a copper wire of cross section `A = 1mm^2` if it carries a current I = 4.5A?`

Answer» Time taken by electron to travel in copper wire is
`t = (i)/(v_d)`
where `v_d = (J)/("ne")`
or `t = (I)/(J//"ne") whre J = i//A`
Hence,
`t = ("ne"AI)/(i)`
`=(10^(28)xx1.6xx10^(-19)xx10^3xx10^(-6))/(4.5) = 3xx10^6s`
Discussion
59.

In a hydrogen discharge tube, the number of protons drifitin across a cross section per second is `1.0xx10^(18)`, while the number of electrons drifting in the opposite direction across the same cross section is `2.7xx 10^(18)` per second. Find the curernt flowing in the tube.

Answer» As electrons and protons are moving in the opposite
directions, they will effectively produce current in the same
direcition and the total current in the tube is
`I = (n_p + n_e)e//t`
`=(1.0xx10^(18) +2.7xx10^(18)) xx1.6xx10^(-19)//1`
`=3.7xx1.6xx10^(-1)A = 0.592 A`
Discussion
60.

A copper wire has a square cross section of 6 mm on a side. The wire is 10 m long and carries a current of 3.6A. The density of free electrons id `8.5x10^(28)//m^3.`Find the magnitude of (a) the current density in the wire , (b) the electric field in the wire. (c ) How much time is required for an electron to travel the length of the wire? `("rh, electrical resistivity, is" 1.72 xx10^(-8) Omegam.)`

Answer» Given ` r= 6 mm, I = 10 m, I = 3.6 A, n = 8.5xx10^(28)//m^3`
(a) To find the currrent density, formula used should be
`J=I//A =(3.6)/((6xx10^(-3))^(2))=(3.6)/(36xx10^(-6))=10^(5) Am^(-2)`
(b) Electric field is
`E = rhoJ = 1.72xx10^(-8) xx10^5 = 1.72 xx10^(-3)Vm^(-1)`
(c ) Time taken is
`t = (I)/(v_d) = (I"ne"A)/(I)`
`=(10xx8.5xx10^(28)xx1.6xx10^(-19)xx(6xx10^(-3))^(2))/(3.6)`
`=1.36xx10^6 s`
Discussion
61.

A cell ofemf E volt with no internal resistance is connected to a wire whose cross section chages. The wire has three sections of equal length. The middle section has a redius a, whereas the radius of the outer two sections is 2a. The raito of the potential differece across section AB to the potential differecne across section CA is A. 5B. 4C. `1//2`D. `1//4`

Answer» Correct Answer - B
in series combination, I is costant, therefore `V propR.`
`(V_(AB))/(V_(CA))=(R_(AB))/(R_(CA))=(rho(I)/(pia^2))/(rho(I)/(pi(2a)^(2)))=(4)/(1)`
Discussion
62.

How many electrons per second pass through a section of wire carrying a current of `0.7A`?

Answer» We known that `I = (q)/(t) = ("ne")/(t) , therefore, n = (it)/(e ) = (0.7xx1)/(1.6x10^(-19)`
So number of electrons in `n = 0.44xx10^(19)`
Discussion