Electrical Fundamentals
Coulombs Law
Learning Objectives:

Define charge and Coulomb's law.

Explain the force of attraction between two charged bodies as a vector quantity.
Electric charge (q), is the extent to which a matter has more or fewer electrons compared to protons.
Electrostatic force that one charge, \(q_{1}\), exerts on another, \(q_{2}\), has a magnitude given by:
\[F = \dfrac{k q_{1}q_{2}}{r^{2}}\]
where r is the distance between \(q_1\) and \(q_2\).
The direction of this force is either toward or away from the other charge depending on whether it is attraction or repulsion.
In the above equation, k is called Coulomb’s constant and its value is given by: \[k = \dfrac{1}{4 \pi \epsilon_0} = 9.0 \times 10^9 N m^2/C^2\] Coulomb’s law in the vector form can be expressed as: \[\vec F_1=k\frac{q_1q_2}{r^2}\hat r_{21}\] where, \(\hat r_{21}\) is a unit vector which is pointing from the other charge to the charge itself.
Solved Example: 18101
The electrical charge associated with one electron, in Coulombs, is:
A. 1
B. 1
C. 6.25 $\times 10^{18}$
D. 1.60 $\times 10^{19}$
The elementary charge, usually denoted as e or sometimes q, is the electric charge carried by a single proton, or equivalently, the magnitude of the electric charge carried by a single electron, which has charge e. This value is constant and equals to ()1.6$ \times 10^{19}$ Coulombs.
Correct Answer: D
Solved Example: 18102
Two identical point charges with mass $m_1$ = $m_2$ = 0.60 grams are suspended in air with strings of length = 250mm. The total angle between them is 15$^\circ$. The charge each of them carrying is:
A. 19.15 nC
B. 27.45 nC
C. 31.87 nC
D. 12.11 nC
Let tension in each string be T.Equating horizontal components $T\sin 7.5^{\circ }=k\dfrac {q_{1}q_{2}}{r^{2}}$
Equating vertical components $T\cos 7.5^{\circ }=mg$
Dividing above two equations, \[\tan 7.5^{\circ }=k\dfrac {q^{2}}{r^{2}mg}\] \[q^{2}=\dfrac {r^{2}mg\tan 7.5^{\circ }}{k}\] where, \[r=2\times 250\times \sin 7.5^{\circ } =65.26\ \mathrm{mm}\] \[q=1.915\times 10^{8}C=19.15\ \mathrm{nC}\]
Correct Answer: A
Solved Example: 18103
Four charges are arranged in a square with sides of length 2.5 cm. The two charges in the top right and bottom left corners are +3.0 $\times$ 10$^{6}$ C. The charges in the other two corners are 3.0 $\times$ 10$^{6}$ C. What is the net force exerted on the charge in the top right corner by the other three charges?
A. 68.7 N
B. 72.9 N
C. 88.2 N
D. 118.0 N
\[F = \dfrac{kq_1q_2}{r^2}\] \begin{align*} F_{14} = F_{12} &= \dfrac{(9\times 10^{9})(3\times 10^{6})(3\times 10^{6})}{(0.025)^2} = 129.5\ N \end{align*}(F$_{12}$ towards q$_2$ and F$_{14}$ towards q$_4$)
\begin{align*} F_{13} &= \dfrac{(9\times 10^{9})(3\times 10^{6})(3\times 10^{6})}{(0.025\sqrt{2})^2}\\ &= 64.7\ N\ (\mathrm{away\ from}\ q_3\ \mathrm{as\ shown}) \end{align*} \begin{align*} \Sigma F_x &= 129.5 + 64.7 \cos 45^\circ = 83.75\ N \end{align*} \begin{align*} \Sigma F_y &= 129.5 + 64.7 \sin 45^\circ = 83.75\ N \end{align*} \begin{align*} F_R &= \sqrt{(\Sigma F_x)^2 + (\Sigma F_y)^2}\\&= \sqrt{(83.75)^2 + (83.75)^2}\\&= 118.43\ N \end{align*}Correct Answer: D
Solved Example: 18104
Three charges 4q, Q and q are in a straight line in the position of 0, $\dfrac{l}{2}$ and l respectively. The resultant force on q will be zero, if Q=? (RSMSSB Lab Asst. 2018)
A. q
B. 2q
C. q$^2$
D. 4q
The force between 4q and q ${{F}_{1}}=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\cdot \dfrac{4q\times q}{{{l}^{2}}}$The force between Q and q ${{F}_{2}}=\dfrac{1}{4\pi {{\varepsilon}_{0}}}\cdot \dfrac{Q\times q}{{{\left(\dfrac{l}{2}\right)}^{2}}}$
Since, the resultant force must be zero, \begin{align*} {{F}_{1}}+{{F}_{2}}&=0\\ \dfrac{4{{q}^{2}}}{{{l}^{2}}}&=\dfrac{4Qq}{{{l}^{2}}}\\ Q&=q \end{align*}
Correct Answer: A
Solved Example: 18105
Two charges of + 4 $\mu$C and 16 $\mu$C are separated from each other by a distance of 0.6 m. At what distance should a third charge of + 6 $\mu$C be placed from + 4 $\mu$C so that no force exerts on it will be zero? (DSSSB JE EE Oct 2019)
A. 0.4 m
B. 0.6 m
C. 1.2 m
D. 0.3 m
Correct Answer: B
Solved Example: 18106
The force between two charges $Q_1$ and $Q_2$ is given by: (NLC GET ECE Nov 2020)
A. Newton's law
B. Coulomb's law
C. Faraday's law
D. Ampere's law
Correct Answer: B
Solved Example: 18107
The force between two charges is 200 N. If the distance between the charges is doubled, the force will be _______. (DFCCIL Jr. Executive S&T Sept 2021)
A. 400 N
B. 100 N
C. 50 N
D. 200 N
Correct Answer: C
Electric Field and Electric Potential
Learning Objectives:

Understand the meaning and significance of electric potential.

Use the electric potential to calculate the electric field.

Use integration to determine electric potential difference between two points on a line, given electric field strength as a function of position along that line.

Calculate how much work is required to move a test charge from one location to another in the field of fixed point charges.

Express in equation form the energy stored in a capacitor.
Electric Field:
Electric field is defined as the electric force per unit charge. Electric field is also known as electrostatic field intensity.
Electric field from a point charge:
The electric field a distance r away from a point charge q is given by:
\[E = \dfrac{k q}{r^2}\]
The electric field from a positive charge points away from the charge; the electric field from a negative charge points toward the charge. Like the electric force, the electric field E is a vector. If the electric field at a particular point is known, the force a charge q experiences when it is placed at that point is given by :
\[F = QE\]
Electric Potential:
Electric potential is more commonly known as voltage. The potential at a point a distance r from a charge q is given by: \[V = \dfrac{k q}{r}\] Electric potential, like potential energy, is a scalar, not a vector.
SI unit for electric potential:
1 Volt = 1 Joule/1 Coulomb or \[V = \dfrac{J}{C}\] (The number of volts indicates the number of joules work or energy per coulomb.)
Potential Difference:
Electric potential difference is the difference in electric potential (V) between the final and the initial location when work is done upon a charge to change its potential energy.
In equation form, the electric potential difference is: \[\Delta V = V_B  V_A = \dfrac{W_{B \rightarrow A}}{q}\] Conventional current flows around a circuit from the positive (+) side of the cell to the negative (). However the electrons are flowing around the circuit in the opposite direction from the negative () side of the cell to the positive (+).
Energy Stored in a Capacitor
\[E = \dfrac{1}{2}CV^2 = \dfrac{QV}{2} = \dfrac{Q^2}{2C}\] where Q is the charge and V the voltage on a capacitor C.
Solved Example: 18201
Energy stored in a condenser of capacity 10 $\mu F$, charged to 6 kV is used to lift a mass of 10g . The height to which the body can be raised is:
A. 180 m
B. 18 m
C. 1.8 m
D. 1800 m
Energy stored in a condenser $E = \dfrac{1}{2} CV^2 = \dfrac{1}{2} \times 10 \times 10^{6} \times 6000^2 = 180\ J$
This will be equal to the potential energy. $180 = mgh, 180 = 0.010 \times 9.81 \times h, h = 1834\ m$
Correct Answer: D
Solved Example: 18202
Two capacitors of capacities 1 $\mu$F and 4 $\mu$F are connected in series with battery of 200 V. The voltage across them are in the ratio of:
A. $\dfrac{1}{2}$
B. $\dfrac{4}{1}$
C. $\dfrac{2}{1}$
D. $\dfrac{1}{4}$
Capacitors in Series will have effective capacitance as:
\[\dfrac{1}{C} = \dfrac{1}{C_1} + \dfrac{1}{C_2}= \dfrac{1}{1} + \dfrac{1}{4}= \dfrac{5}{4}\]
\[C = \dfrac{4}{5} = 0.8\ \mu F\]
Charge on the effective capacitor is given by, $Q = CV = 0.8 \times 10^{6} \times 200 = 160 \times 10^{6}\ \mathrm{Coulombs}$
Capacitors in series will have same charge as that of equivalent. $Q = Q_1 = Q_2 = 160 \times 10^{6}\ \mathrm{Coulombs}$
\[V_1 = \dfrac{Q_1}{C_1}
= \dfrac{160 \times 10^{6}}{1 \times 10^{6}}
= 160\ V\]
\[V_2 = \dfrac{Q_2}{C_2}
= \dfrac{160 \times 10^{6}}{4 \times 10^{6}}
= 40\ V\]
\[\dfrac{V_1}{V_2} = \dfrac{160}{40} = \dfrac{4}{1}\]
Correct Answer: B
Solved Example: 18203
How do you arrange four equal capacitors of 4 $\mu F$ to get effective capacity of 3 $\mu F$ ?
A. Three in series, one in parallel
B. Two in parallel, two in series
C. Three in parallel, one in series
D. All four in series
If three capacitors are in series, then effective capacitance = 12 $\mu$ F. This is in series with 4 $\mu$ F. So effective capacitance $\dfrac {1}{C_{\mathrm{eff}}}=\dfrac {1}{12}+\dfrac {1}{4}=\dfrac {1+3}{12}=\dfrac {4}{12}, \quad C_{\mathrm{eff}}=\dfrac {12}{4}=3\mu F$
Correct Answer: C
Solved Example: 18204
In an experiment, a charged drop of oil of mass 1.8 $\times 10^{14}$ kg is stationary between the plates of a parallel plate condenser separated by a distance of 9 mm and charged to a potential difference of 2000 V. The charge on the drop is:
A. 8 $\times 10^{19}$ C
B. 6 $\times 10^{19}$ C
C. 5 $\times 10^{19}$ C
D. 4 $\times 10^{19}$ C
Vertical Force = qE
But
\[E = \dfrac{V}{d}
= \dfrac{2 \times 10^3}{9 \times 10^{3}}
= 0.22 \times 10^6 V/m\]
Also, qE = mg
\[q = \dfrac{mg}{E}
=\dfrac{1.8 \times 10^{14} \times 9.8}{0.22 \times 10^6}
=8.02 \times 10^{19} C\]
Correct Answer: A
Solved Example: 18205
Point charge q moves from point P to point S along the path PQRS (figure shown) in a uniform electric field E pointing coparallel to the positive direction of the Xaxis. The coordinates of the points P,Q,R and S are (a,b), (2a,0), (a,b) and (0,0) respectively. The work done by the field in the above:
A. qEa
B. qE$\sqrt{(2a)^2+b^2}$
C. qEa$\sqrt{2}$
D. qEa
As electric field is a conservative field Hence the work done does not depend on path. Instead of following red path, we can follow blue path as shown in the figure, which is much simpler.
\begin{align*} W_{PQRS} &= W_{POS} = W_{PO} + W_{OS}\\ &=Fb \cos 90^\circ +Fa \cos180^\circ\\ &=0 + qEa(1)\\ &=qEa \end{align*}Correct Answer: A
Solved Example: 18206
A capacitor is charged by using a battery which is then disconnected. A dielectric slab is then slipped between the plates, which results in:
A. Reduction of charge on the plates and increase of potential difference across the plates.
B. Increase in the potential difference across the plate, reduction in stored energy, but no change in the charge on the plates.
C. Decrease in the potential difference across the plates, reduction in the stored energy, but no change in the charge on the plates.
D. None of the above.
Battery in disconnected so Q will be constant as C $\propto$ K. So with introduction of dielectric slab capacitance will increase using Q = CV, V will decrease and using U= $\dfrac{Q^2}{2C}$, energy will decrease.
Correct Answer: C
Electric Current
Learning Objectives:

Define electric current.

Calculate electric current from charge and time.
Electric current is defined as the rate at which charge flows through a surface (such as cross section of a wire).
\[di(t) = \dfrac{dq(t)}{dt}\]
Or in stedy state format, \[I = \dfrac{Q}{t}\] I = electric current rate at which charge flows in a certain pathway
Q = amount of charge flowing past a certain point
t = time
SI unit for electric current:
1 Ampere = 1 Coulomb/1 second
Solved Example: 18301
Rate of flow of charge through crosssectional area is known as: (MP Aug 2017 Shift I)
A. Current
B. Voltage
C. Electric potential
D. Charge velocity
Correct Answer: A
Solved Example: 18302
Current flowing through a conductor, when 2 $\times$ 10$^7$ electrons pass in 1 $\mu$sec will be:
A. 1.6 $\times$ 10$^{6}$
B. 3.2 $\times$ 10$^{6}$
C. 4.2 $\times$ 10$^{6}$
D. 6 $\times$ 10$^{6}$
Each electron has a charge of 1.6 $\times$ 10$^{19}$ C.
\begin{align*}
\mathrm{current} = \dfrac{\mathrm{charge}}{\mathrm{time}}
&= \dfrac{1.6 \times 10^{19} \times 2 \times 10^7}{1 \times 10^{6}}\\
&= 3.2 \times 10^{6}\ \mathrm{A}
\end{align*}
Correct Answer: B
Resistance
Learning Objectives:

Define resistance (R) and know how it can be altered.

Use Ohm’s law to design circuits.

Calculate resistance from resistivity (conductivity)
Electrical Resistance:
Electrical resistance is a measure of the difficulty to pass an electric current through that conductor. \[R = \dfrac{V}{I}\] or more commonly written as: \[V = IR\](known as Ohm’s Law), where R = electric resistance (this is the resistance to flow of charge through a device).
An object or device with greater resistance will require a greater voltage to produce a certain amount of current.
SI unit for resistance is Ohm:
1 Ohm = 1 Volt/1 Ampere or W = \(\dfrac{V}{A}\)
(The number of ohms indicates how many volts are required to produce 1 Ampere of current.)
Resistivity:
Resistivity is a measure of the resisting power of a specified material to the flow of an electric current.
For a conductor of length L, electrical resistivity \(\rho\), and crosssectional area A, the resistance is: \[R = \dfrac{\rho L}{A}\]
Internal Resistance:
Internal resistance is the resistance within a battery, or other voltage source, that causes a drop in the source voltage when there is a current. Thus, the voltage \(V\) of the battery is related to its emf \(E\) and internal resistance \(r\) via \[V = {E}  I\,r.\]
Solved Example: 18401
Internal resistance of cell when there is current of 0.40 A when a battery of 6.0 V is connected to a resistor of 13.5 $\Omega$ is:
A. 1.5 $\Omega$
B. 2.3 $\Omega$
C. 3.5 $\Omega$
D. 4.5 $\Omega$
Total resistance of the circuit, including internal resistance, can be calculated by:
\[R = \dfrac{V}{I}
= \dfrac{6}{0.4}
= 15\ \Omega\]
This resistance is including internal resistance. Out of this 15 $\Omega$ , 13.5 $\Omega$ is external resistance.
Internal resistance,
\[ r = R_{Total}  R_{External} = 15  13.5 = 1.5\ \Omega\]
Correct Answer: A
Solved Example: 18402
Maximum current a battery of e.m.f. 3.0 V and internal resistance 1.0 $\Omega$ is:
A. 4.0 A
B. 5.0 A
C. 3.0 A
D. 30 A
Maximum current will flow, when the ends of the battery are shortcircuited, i.e. the external resistance will be zero.
In such case, the electrons will encounter only internal resistance of the battery.
\[i = \dfrac{V}{R}
= \dfrac{3}{1}
= 3\ A\]
Correct Answer: C
Solved Example: 18403
The resistance of an electric bulb drawing 1.2 A current at 6.0 V is ______________.
A. 0.5 $\Omega$
B. 5 $\Omega$
C. 0.2 $\Omega$
D. 2 $\Omega$
\[R = \dfrac{V}{I} = \dfrac{6}{1.2} = 5\ \Omega\]
Correct Answer: B
Solved Example: 18404
The unit of resistivity is _________________.
A. Ohm
B. Ohm / m
C. Ohmm
D. Mho
Correct Answer: C
Solved Example: 18405
Specific resistance of a wire can be measured by formula:
A. $\dfrac{R}{L}$
B. $\dfrac{RA}{L}$
C. $\dfrac{RL}{A}$
D. $\dfrac{A}{RL}$
Resistance of a wire depends as follows:
Directly proportional to the length of the wire,
Inversely proportional to the crosssectional area of the wire
\[R \propto \dfrac{L}{A} = \rho \cdot \dfrac{L}{A}\] Hence, \[\rho = \dfrac {RA}{L}\]
Correct Answer: B
Solved Example: 18406
Current of 0.75 A, when a battery of 1.5 V is connected to wire of 5 m having cross sectional area 2.5 $\times$ 10$^{7}$ m$^2$, will have resistivity, in $\Omega$.m:
A. 1 $\times$ 10$^{7}$
B. 1.1 $\times$ 10$^{7}$
C. 2 $\times$ 10$^{7}$
D. 2.1 $\times$ 10$^{7}$
\[R = \dfrac{V}{I} = \dfrac{1.5}{0.75} = 2\ \Omega\] Also, \begin{align*} R &= \dfrac{\rho \cdot L }{ A}\\ 2 &= \dfrac{\rho \cdot 5}{2.5 \times 10^{7}}\\ \rho &= \dfrac{2 \times 2.5 \times 10^{7}}{5} = 1 \times 10^{7}\ \Omega.m \end{align*}
Correct Answer: A
Solved Example: 18407
Resistivity of a wire depends on:
A. Length
B. Material
C. Cross section area
D. None of the above
textbfResistance of a wire depends upon three factors.
The length of the wire,
The crosssectional area of the wire, and
The resistivity of the material composing the wire.
Resistivity should not be confused with
resistance of a wire.
Resistivity is a specific property of a material. It depends only upon
the material, and NOT on other geometrical factors such as
length or crosssectional area.
Correct Answer: B
Electric Power
Learning Objectives:

Define the concept of power and understand the relationship between current, voltage, and power.

Calculate power in electric circuits (both electrical energy transfer and energy lost through resistance)

Understand heating effect of electric current.
Electric power is the rate, per unit time, at which electrical energy is transferred by an electric circuit.
The SI unit of power is the Watt, one Joule per Second.
\[P = V I = R I^2 = \dfrac{V^2}{R} = \dfrac{VQ}{t}\]
Solved Example: 18501
If we double both current and voltage keeping resistance constant then power is:
A. Unchanged
B. Halves
C. Doubles
D. Quadruples
\[P = V_{new} \cdot I_{new} = 2V_{old} \cdot 2I_{old} = 4P_{old}\]
Correct Answer: D
Solved Example: 18502
If a light bulb is switched on for 40 s and it consumes 2400 J of electrical energy then its power is:
A. 60 W
B. 70 W
C. 80 W
D. 90 W
$P = \dfrac{W}{t} = \dfrac{2400}{40} = 60\ W$
Correct Answer: A
Solved Example: 18503
Amount of energy supplied by current in unit time is called:
A. Electric energy
B. Electric power
C. Resistance
D. Friction
Rate of energy is power. If the power is given by displacement of body, it is mechanical power. If it is given by movement of electrons (in the form of current), it is electrical power.
Correct Answer: B
Solved Example: 18504
If resistance of an electric bulb is 500 $\Omega$ and voltage across its ends is 250 V then power consumed by it is:
A. 130 W
B. 125 W
C. 120 W
D. 200 W
$P = V \cdot I = V \cdot \left(\dfrac{V}{R}\right) = \dfrac{V^2}{R} = \dfrac{250^2}{500} = 125\ W$
Correct Answer: B
Magnetostatics
Learning Objectives:

Define the concept of magnetism.

Understand fundamental theories of electricity and magnetism.

Be able to calculate electromagnetic forces and fields.

Calculate power in electric circuits (both electrical energy transfer and energy lost through resistance)

Understand heating effect of electric current.
Terminology of Magnetostatics:
A magnet is as solid material that possesses the property of attracting the iron pieces. When a magnet is rolled into iron pieces it will attract the iron pieces towards the end points.
The points at which the iron pieces accumulate maximum are called poles of the magnet while imaginary lines joining these poles called axis of the magnet.
The natural magnets have the property of magnetism naturally present whereas the electromagnets are formed by passing an electric current around a certain material. The material then acts as magnet as long as the current is present. But it loses its magnetic properties as soon as the current stops.
Magnetic Field: The magnetic field is defined as the region near a magnet within which the effect or influence of the magnet is felt.
Magnetic Lines of Force: A line of force is defined as a line along which an isolated Npole would travel if it is allowed to move freely in a magnetic field.
Properties of Magnetic Lines of Forces:

The magnetic lines of force always form a closed loop. They originate from N pole and terminate in Spole.

The magnetic lines of force do not cross or intersect each other.

The magnetic lines of force which are parallel to each other and are acting in the same direction tend to repel each other.

The magnetic lines of force behave like stretched elastic band and always try to contract in length.

The magnetic lines of force always try to follow the minimum opposition (reluctance) path.
Magnetic Circuit:
A magnetic circuit is defined as the closed path followed by the magnetic lines of force i.e. flux. This is very similar to an electric circuit which states that the electric circuit is the closed path provided for the electric circuit. The quantities associated with magnetic circuits are MMF, flux, reluctance permeability etc.
Magnetic Flux (\(\phi\)):
The magnetic flux is defined as the total number of magnetic lines of force in a magnetic field. It is denoted by ‘\(\phi\)’ and is measured in Webber (Wb). One Webber is defined as the flux radiated out by a unit Npole.
1Wb = 10\(^8\) lines of force
Magnetic Flux Density (B):
The flux per unit area (A), measured in a plane perpendicular to the flux is defined as the flux density. It is measured in tesla (T) or (Wb/m\(^2\)).
\[B = \dfrac{\phi}{A}\]
Magnetic Field Strength or Magnetic Field Intensity (H):
The magnetic field strength at a point in the magnetic field is defined as the force experienced by a unit North Pole placed at that point in the magnetic field. It is measured in Newton per Weber (N/Wb).
Magneto Motive Force (MMF):
The Magneto Motive Force is defined as the force responsible for the generation of the flux. It is nothing but the work done on a unit magnetic pole to take it around a closed magnetic circuit.
The MMF is the driving force behind the magnetic circuit. It is given by the product of the number of turns of the magnetizing coil and the current passing through the coil. Mathematically, \[MMF= N.I\] Where,
N = number of turns of the magnetizing coil
I = current passing through the coil
The SI unit of the MMF is Ampereturn (AT)
Reluctance (S):
The reluctance of a material is the opposition offered by that material for passage of magnetic lines of force (magnetic flux) through it. It is denoted by ‘S’. The reluctance of a material is directly proportional to the length of the magnetic circuit and inversely proportional to the area of cross section. \[S = \dfrac{MMF}{\phi}\] \[S = \dfrac{NI}{\phi}\] Where,
K = constant of proportionality = reciprocal of the absolute permeability \[S = \dfrac{l}{\mu a}\] It is measured in Webber/ ampere (Wb/A). Also the reluctance of a material is defined as the ratio of magneto motive force to the magnetic flux produced. \[S \propto \dfrac{l}{a}\] \[S = \dfrac{Kl}{a}\] \[d\textbf{H} = \dfrac{I}{4 \pi} \dfrac{d\textbf{l} \times \hat{\textbf{R}}}{R^2} (A/m)\] Over a conductor path l, we obtain H as the line integral, \[\textbf{H} = \dfrac{I}{4 \pi} \int_l \dfrac{d\textbf{l} \times \hat{\textbf{R}}}{R^2} (A/m)\]
Solved Example: 18601
Which, among the following qualities, is not affected by the magnetic field?
A. Moving charge
B. Change in magnetic flux
C. Current flowing in a conductor
D. Stationary charge
A stationary charge is not affected by a magnetic field because stationary charges do not have any velocity. Magnetic field cannot occur in a particle having zero velocity.
Correct Answer: D
Solved Example: 18602
Magnetomotive force is equal to:
A. Current / number of turns per unit length
B. Current * number of turns per unit length
C. Current * number of turns
D. Current / number of turns
MMF is ability to produce flux = product of current flowing and number of turns.
Correct Answer: C
Solved Example: 18603
The relation between the direction of current and the direction of magnetic field is?
A. Same direction
B. Opposite direction
C. Perpendicular
D. Unrelated
When a conductor carries current, the force developed in the conductor, the current in the conductor and the magnetic field in the conductor are mutually perpendicular to each other.
Correct Answer: C
Solved Example: 18604
Calculate the magnetic flux density, in T, if the magnetic field strength is 2 A/m.
A. $4 \pi \times 10^{7}$
B. $8 \pi \times 10^{7}$
C. $10 \pi \times 10^{7}$
D. $12 \pi \times 10^{7}$
We know that: $\mu_0=\dfrac{B}{H}$ Substituting H from the question, we get B = $8 \pi \times 10^{7}$ T.
Correct Answer: B
Solved Example: 18605
Calculate the magnetic field strength if the magnetic flux density is 4 $\times \pi \times$ T.
A. $10^{7}/16*\pi^2$A/m
B. $10^{7}$A/m
C. $10^7$A/m
D. $10^{7}$A
We know that: $\mu_0=\dfrac{B}{H}$. H = $10^7$ A/m.
Correct Answer: C
Solved Example: 18606
Find H = ___________ A/m at the center of a circular coil of diameter 1 m and carrying a current of 2 A. (SSC JE EE Sep 2019 Morning)
A. 0.6366
B. 0.1636
C. 6.366
D. 2
Correct Answer: D
Solved Example: 18607
Calculate the flux density at a distance of 5 cm from a long straight circular conductor carrying a current of 250 A and placed in air. (SSC JE EE Dec 2020)
A. 10$^2$ Wb/m$^2$
B. 10$^{2}$ Wb/m$^2$
C. 10$^{3}$ Wb/m$^2$
D. 10$^3$ Wb/m$^2$
Correct Answer: C
Solved Example: 18608
Two identical coils A and B of 1000 turns each lie in parallel plane such that 80% of the flux produced by one coil links with the other. If a current of 5 A flowing in A produces a flux of 0.05 mWb, then the flux linking with coil B is: (SSC JE EE Sep 2019 Morning)
A. 0.4 mWb
B. 0.04 mWb
C. 4 mWb
D. 0.004 mWb
Correct Answer: B
Solved Example: 18609
The S.I. unit magnetic permeance is: (SSC JE EE Jan 2018 Evening)
A. Henry
B. Weber
C. Tesla
D. Coulomb
Correct Answer: A
Solved Example: 18610
Magnetic flux will be _________ if the surface area vector of a surface is perpendicular to the magnetic field. (UPPCL JE Nov 2016)
A. Zero
B. Unity
C. Close to maximum
D. Maximum
Correct Answer: A
Solved Example: 18611
A coil of 360 turns is linked by a flux of 200 $\mu$ Wb. If the flux is reversed in 0.01 second, then find the EMF induced in the coil. (SSC JE EE Sep 2019 Evening)
A. 7.2 V
B. 0.72 V
C. 14.4 V
D. 144 V
Correct Answer: C
Solved Example: 18612
Two magnetic poles are located 5 cm apart in air. If each pole has a strength of 5 mWb, find the force of repulsion between them. (SSC JE EE Dec 2020)
A. $\dfrac{1}{{{\pi ^2}}}N$
B. $\dfrac{6250}{{{\pi ^2}}}N$
C. $\dfrac{625}{{{\pi ^2}}}N$
D. $\dfrac{62.5}{{{\pi ^2}}}N$
Correct Answer: B
Solved Example: 18613
A long straight circular conductor placed in air is carrying a current of 250 A. Find the magnetising force at a distance of 5 cm from the conductor. (SSC JE EE March 2021 Morning)
A. $\dfrac{{5000}}{\pi }AT/m$
B. $\dfrac{{500}}{\pi }AT/m$
C. $2500 AT/m$
D. $\dfrac{{2500}}{\pi }AT/m$
Correct Answer: D
Solved Example: 18614
The force on the currentcarrying conductor in a magnetic field depends upon:
(a) the flux density of the field
(b) The strength of the current
(c) The length of the conductor perpendicular to the magnetic field
(d) The directions of the field and the current
(UPPCL AE EE Nov 2019 Shift II)A. (a), (c) and (d) only
B. (a), (b)and (c) only
C. (a), (b) and (d) only
D. (a), (b), (c) and (d)
Correct Answer: D