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Electricity and Electric Current

Q: A long straight wire of radius a carries a steady current \$I.\$ The current is uniformly distributed over its cross-section. The ratio o the magnetic fields \$B\$ and \$B',\$ at radial distances \$\dfrac{a}{2}\$ and \$2a\$ respectively, from the axis of the wire is

Consider two amperian loops of radius \$a/2\$ and \$2a\$ as shown in the diagram\$.\$Applying ampere's circuital law for these loops\$,\$ we get\$\oint {B.dL = } {\mu _0}{I_{enclosed}}\$For the smaller loop\$,\$\$ \Rightarrow B \times 2\pi \dfrac{a}{2} = {\mu _0} \times \dfrac{1}{{\pi {a^2}}} \times {\left( {\dfrac{a}{2}} \right)^2}\$\$ = {\mu _0}I \times \dfrac{1}{4} = \dfrac{{{\mu _0}I}}{4}\$\$ \Rightarrow {B_1} = \dfrac{{{\mu _0}I}}{{4\pi a}}\$\$B' \times 2\pi \left( {2a} \right) = {\mu _0}I\$\$\dfrac{B}{{B'}} = \dfrac{{{\mu _0}I}}{{4\pi a}} \times \dfrac{{4\pi a}}{{{\mu _0}I}} = 1\$Hence,option \$(C)\$ is correct answer.

Q: A current of 1 A is drawn by a filament of an electric bulb. Number of electrons passing through a cross section of the filament in 16 seconds would be roughly:

\$\textbf{Given : } I= 1A, t=16s, e= 1.6\times 10^{-19} C\$\$\textbf {Step 1: Current drawn by the filament} \$ \$I= \dfrac{Q}{t} = \dfrac{Ne}{t}\$ (Since \$Q= Ne\$, where N is the number of electrons crossing in time t)\$\Rightarrow N= \dfrac{I \ t}{e}\$ \$....(1)\$\$\textbf {Step 2: Substituting the values in equation (1)} \$\$\ N=\dfrac { I\ t }{ e } =\dfrac { 1A\times 16 s} { 1.6\times { 10 }^{ -19 } C} ={ 10 }^{ 20 }\$Hence, number of electrons passing through a cross section of the filament in 16 seconds would be \${ 10 }^{ 20 }\$.

Q: Calculate the number of electrons constitute one coulomb of charge.

\$q=ne\$where, \$q\rightarrow\$ Change \$, n\rightarrow \$no. of \$e^-\$ \$e\rightarrow \$ Change on \$e^-\$\$\Rightarrow 1c=n\left( 1.6\times 10^{-19}\right)\$\$\Rightarrow n=\dfrac{10^{19}}{1.6}\$\$\Rightarrow n=625.10^{16}\$ electronsHence, the answer is \$625.10^{16}\$ electrons.

Q: If the charge on an electron is \$1.6 \times 10^{-19}\$ coulombs , how many electrons should pass through a conductor in \$1\$ second to constitute \$1\$ ampere current ?

We know that \$I = \dfrac {Q}{t}\$\$\Rightarrow 1 \,A = \dfrac {Q}{1 \,s}\$\$\Rightarrow Q = 1 \,C\$Now, charge of one electron is \$1.6 \times 10^{-19}\$ coulombsTherefore, in \$1\$ coulomb charge, number of electrons \$=\$\$\dfrac {1}{1.6 \times 10^{-19}} = 0.625 \times 10^{19} \$\$= 6.25 \times 10^{18}\$ electrons

Q: Charge on an electron is \$1.6 \times 10^{-19}\$ coulomb. Number of electrons passing through the wire per second on flowing of 1 ampere current through the wire will be

The current is defined as rate of flow of charged particle.i.e.i \$ = \dfrac{Q}{t} \$Hence the charge flown is,Q \$ = i \ t \$Hence Q \$ = 1.0 \times 1 = 1 C \$The charge Q is due to electrons, therefore,Q \$ = N\times e \$Here Q \$ = 1 C \$If N is number of electrons passing and e is the electronic charge of value \$ 1.6 \times {10}^ {-19} \$Then,N \$ = \dfrac {Q}{e} = \dfrac {1}{1.6 \times {10}^ {-19} } = 0.625 \times {10}^ {19} electrons \$

Q: A current flowing through a wire depends on time as \$I=3t^2+2t+5\$. The charge flowing through the cross section of the wire in time from \$t=0\$ to \$t=2sec \$ is

\$\textbf{Step 1: Current through the wire}\$We know that,\$I=\dfrac{dq}{dt}\$\$\Rightarrow dq= I \ dt\$ \$....(1)\$\$\textbf{Step 2: Integrating the equation (1) with proper limits}\$Time interval: \$t=0s\$ to \$t=2s\$\$ \int{dq}=\int_{0}^{2}{Idt}\$\$I=3t^2+2t+5\$\$\Rightarrow q=\int_{0}^{2}{(3t^2+2t+5)dt}=[t^3+t^2+5t]_{0}^{2}=(8+4+10)=22C\$Hence Option \$(A)\$ is correct.

Q: A spherical conducting shell of inner radius r1 and outer radius r2 has a charge Q. A charge q is placed at the centre of the cell.(a) What is the surface charge density on the (i) inner surface (ii) outer surface of the shell.(b) Write the expression for electric field at a point x r2 from the centre of the shell.

We have,\$radii=r_1,r_2\$\$charge=Q,q\$So,a) Surface charge density on the inner surface is:\$\dfrac{q}{4\pi r_1^2}\$Surface charge density on the outer surface is:\$\dfrac{q+Q}{4\pi r_2^2}\$from b) For finding the electric field at the distance x from the center is:For this Apply the Gauss law,\$E\times4\pi x^2=4\pi(Q+q)\$\$E=\dfrac{Q+q}{x^2}\$

Q: The current in a conductor varies with time t as \$I=2t+3t^{2}\$ amp. where I is ampere and t in second. Electric charge flowing through a section of the conductor during \$t=2\:sec\$ to \$t=3\:sec\$ is:-

\$\textbf{Step 1: Electric charge passing through the conductor} \$ We know, Instantneous current across a conductor can be defined as:\$I = \dfrac{dQ}{dt} \$ \$\Rightarrow dQ = I dt\$ \$\Rightarrow dQ = (2t + 3t^2) dt \$ \$....(1)\$\$\textbf{Step 2: Integrate the above equation } \$Electric Charge flowing through section of the conductor during \$t = 2s \$ to \$t = 3s \$ can be obtained by integrating the equation \$(1)\$\$\displaystyle \int dQ = \int^{t = 3s}_{t = 2s} (2t + 3t^2) dt \$\$\Rightarrow \Delta Q = 2 \left [\dfrac{t^2}{2} \right ]^{3}_{2} + 3 \left [\dfrac{t^3}{3} \right ]^{3}_{2} \$ \$\Rightarrow \Delta Q = [9 - 4] + [27 - 8] \$\$ = 24 C \$ Hence, Option B is correct

Q: A current of \$4.8\ A\$ is flowing in a conductor. The number of electrons passing per second through the conductor will be :

Given,\$ e= 1.6 \times 10^{-19}\ C \$, \$ I=4.8\ A \$Let number of electrons per second be \$n\$.The current is defined as rate of flow of charge,\$I = \dfrac{q}{t} \$ Hence\$q = n \times e \$Therefore, \$I = \dfrac{n\times e}{t}\$Substituting values,4.8 \$ = n \times 1.6\times 10^{-19} \$Hence, \$n = \dfrac {4.8}{1.6\times 10^{-19}} = 3\times 10^{19} \$ electrons per second.

Q: A current of \$5 A\$ flows through a conductor. Calculate the number of electrons flowing through the conductor in \$2s.\$ [Take charge of electron as \$1.6 \times10^{-19}\$

Total charge that flows in 2s is given by \$q=It=5\times 2=10C\$Now, charge on one electron is given by \$e=1.6\times 10^{-19}C\$\$N=\$ number of electrons\$q=Ne\$\$\implies N=\dfrac{q}{e}\$\$\implies N=\dfrac{10}{1.6\times 10^{-19}}\$\$\implies N=6.25\times 10^{19}\$Answer-(B)

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A long straight wire of radius a carries a steady current I. The current is uniformly distributed over its cross-section. The ratio o the magnetic fields B and B′, at radial distances 2a​ and 2a respectively, from the axis of the wire is

A current of 1 A is drawn by a filament of an electric bulb. Number of electrons passing through a cross section of the filament in 16 seconds would be roughly:

Calculate the number of electrons constitute one coulomb of charge.

If the charge on an electron is 1.6×10−19 coulombs , how many electrons should pass through a conductor in 1 second to constitute 1 ampere current ?

Charge on an electron is 1.6×10−19 coulomb. Number of electrons passing through the wire per second on flowing of 1 ampere current through the wire will be

A current flowing through a wire depends on time as I=3t2+2t+5. The charge flowing through the cross section of the wire in time from t=0 to t=2sec is

The current in a conductor varies with time t as I=2t+3t2 amp. where I is ampere and t in second. Electric charge flowing through a section of the conductor during t=2sec to t=3sec is:-

A current of 4.8 A is flowing in a conductor. The number of electrons passing per second through the conductor will be :

A current of 5A flows through a conductor. Calculate the number of electrons flowing through the conductor in 2s. [Take charge of electron as 1.6×10−19

A long straight wire of radius a carries a steady current $I .$ The current is uniformly distributed over its cross-section. The ratio o the magnetic fields $B$ and $B^{'},$ at radial distances $\frac{a}{2}$ and $2 a$ respectively, from the axis of the wire is

If the charge on an electron is $1.6 \times 1 0^{- 19}$ coulombs , how many electrons should pass through a conductor in $1$ second to constitute $1$ ampere current ?

Charge on an electron is $1.6 \times 1 0^{- 19}$ coulomb. Number of electrons passing through the wire per second on flowing of 1 ampere current through the wire will be

A current flowing through a wire depends on time as $I = 3 t^{2} + 2 t + 5$ . The charge flowing through the cross section of the wire in time from $t = 0$ to $t = 2 s e c$ is

The current in a conductor varies with time t as $I = 2 t + 3 t^{2}$ amp. where I is ampere and t in second. Electric charge flowing through a section of the conductor during $t = 2 s e c$ to $t = 3 s e c$ is:-

A current of $4.8 A$ is flowing in a conductor. The number of electrons passing per second through the conductor will be :

A current of $5 A$ flows through a conductor. Calculate the number of electrons flowing through the conductor in $2 s .$ [Take charge of electron as $1.6 \times 1 0^{- 19}$