Consider the following statements:
The force per unit length between two stationary parallel wires carrying (steady) currents _____.
A. is inversely proportional to the separation of wires
B. is proportional to the magnitude of each current
C. satisfies Newton's third law
Out of this _____.The force between two current-carrying parallel conductors:
\(\frac{F}{l}=~\frac{{{\mu }_{0}}}{2\pi }\frac{{{i}_{1}}{{i}_{2}}}{d}\)
It satisfies newton’s third law.Concept:
We know the force per unit length between the two conductors is
\(\frac{F}{l} = \frac{{{μ _0}{I_1}{I_2}}}{{2π d}}\)
Calculation:
Here d = distance = 10 × 10^{-3} meter
μ0 = 4π × 10-7F/m
And currents I_{1} = I_{2} = 150 Ampere
\(\therefore \frac{F}{l} = \frac{{4π × {{10}^{ - 7}} × 150 × 150}}{{2π × 10 × {{10}^{ - 3}}}} = 0.45\;N/m\)
Magnetic circuits are analogous to electric circuits. Some of the quantities are given below.
Magnetic Circuit |
Electric Circuit |
Flux |
Current |
Permeance |
Conductance |
Reluctance |
Resistance |
Permeability |
Conductivity |
Flux density |
Current density |
Magnetic field |
Electric field |
Magnetomotive force |
Electromotive force |
Gauss' law states that the electric flux through any closed surface is equal to the total charge inside divided by ε_{0}. Charges are the source and sinks of the electric field.
The field outside a uniformly distributed spherically symmetric charge distribution looks like the field of a point charge. If Q is positive, the field points outward, and if Q is negative, it points inward. Q_{inside }can be written as the charge density ρ = Q/V times the volume of the charged sphere.
We can therefore also write \(E = \frac{{\rho {R^3}}}{{3{\epsilon_0}{r^2}}}\)
The field inside the uniformly distributed spherically symmetric charge distribution increases linearly with r. Its direction is outward for a positive distribution, and inward for a negative distribution.
If E is zero, the charge density in the ideal conductor is zero.
The total electric field at any point is equal to the vector sum of the separate electric fields that each point charge would create in the absence of the others. That is,
\(\;\mathop \sum \limits_i {E_i} = {E_1} + {E_2} + {E_3} + \ldots\)
The electric field is nothing but the potential gradient of that particular point.CONCEPT:
Coulomb’s law: When two charged particles of charges q1 and q2 are separated by a distance r from each other then the electrostatic force between them is directly proportional to the multiplication of charges of two particles and inversely proportional to the square of the distance between them.
Force (F) ∝ q1 × q2
\(F \propto \frac{1}{{{r^2}}}\)
\(F = K\frac{{{q_1} \times {q_2}}}{{{r^2}}}\)
Where K is a constant = 9 × 109 Nm2/C2
Calculations:
Consider new charge + 4μC is placed d m apart from old +4 μC charge and (x + 0.6) m apart from -16 μC charge.
Let,
q_{A} = + 4 μC at point A
q_{B} = - 16 μC at point B
q_{C} = + 6 μC at point C
Since, net force on charge q_{C} will zero.
∴ |F_{CA}| = |F_{CB}|
From above concept,
\(\frac{Kq_Aq_C}{d^2}=\frac{Kq_Bq_C}{(0.6+d)^2}\)
\(\frac{24}{d^2}=\frac{96}{(0.6+d)^2}\)
4d^{2} = (0.6 + d)^{2}
4d^{2} = 0.36 + d^{2} + 1.2d
3d^{2} - 1.2d - 0.36 = 0
d_{1} = - 0.2 m
d_{2} = + 0.6 m
Hence, according to option + 0.6 m distance should a third charge of + 6 μC be placed from + 4 μC so that no force exerts on it will be zero.
We know that
\(E = \frac{Q}{{4\pi \epsilon{{\rm{r}}^2}}}\)
E = electrical field
Q = charge
r = distance
∴ the electric field strength of a charge decreases with square of distance.What is the absolute permittivity of a substance?
Absolute permittivity:
Term |
Point Charge |
Line charge density |
Surface charge density |
Volume charge density |
Denoted by |
Q |
\({\rho _L}\) |
\({\rho _s}\) |
\({\rho _v}\) |
Unit |
Coulomb or C |
Coulomb / meter or (C/m) |
Coulomb / meter^{2} or (C/m^{2}) |
Coulomb / meter^{3} or (C/m^{3}) |
Net charge(Q) |
Q |
\(\mathop \smallint \nolimits_l {\rho _L}dl\) |
\(\mathop \smallint \nolimits_s {\rho _s}ds\) |
\(\mathop \smallint \nolimits_v {\rho _v}dv\) |
Electric field intensity (E) |
\(\frac{1}{{4\pi {\varepsilon _0}}}\smallint \frac{Q}{{{R^2}}}{a_R}\) |
\(\frac{1}{{4\pi {\varepsilon _0}}}\mathop \smallint \nolimits_l \frac{{{\rho _L}dl}}{{{R^2}}}{a_R}\) |
\(\frac{1}{{4\pi {\varepsilon _0}}}\mathop \smallint \nolimits_s \frac{{{\rho _s}ds}}{{{R^2}}}{a_R}\) |
\(\frac{1}{{4\pi {\varepsilon _0}}}\mathop \smallint \nolimits_v \frac{{{\rho _v}dv}}{{{R^2}}}{a_R}\) |
Option 2 is correct, i.e. Conductance.
Conductance:
Note:
Conductivity:
Resistivity (ρ):
Resistance:
It is the magnetic field lines that form a closed loop as shown:
CONCEPT:
Force (F) ∝ q1 × q2
\(F \propto \frac{1}{{{r^2}}}\)
\(F = K\frac{{{q_1} \times {q_2}}}{{{r^2}}}\)
Where K is a constant = \(\frac{1}{{4\pi {\epsilon_0}}}\)= 9 × 109 Nm2/C2
CALCULATION:
The force of attraction between two charges is given by,
\(F = \frac{1}{{4\pi {\epsilon_0}}}\frac{{\left| {{q_1}{q_2}} \right|}}{{{r^2}}}\;\)
Given that, q_{1} = 1 C, q_{2} = -1 C, r = 1 m
\(F = \frac{1}{{4\pi {\epsilon_0}}} \times 1 = 9 \times {10^9}N\)
So option 1 is correct.
Under the static condition, the electric field inside the solid perfect conductor is zero in electrostatic equilibrium even if:
Now, it is known that a perfect solid conductor is an equipotential body. The potential inside the conductor is the same as the potential at the surface
Since the surface of the solid conductor is an equipotential surface hence the electric field will be perpendicular or in other words normal to the perfectly solid conductor surface.
Which of the following law states that “The total flux out of a closed surface is equal to the net charge within the surface”?
Gauss law for electric field:
Gauss's law states that the net flux of an electric field in a closed surface is directly proportional to the enclosed electric charge.
The net flux of a given electric field through a given surface, divided by the enclosed charge should be equal to a constant.
Integral form: \(\phi = \smallint E.nds = \frac{Q}{{{\varepsilon _0}}}\)
\(\smallint D.nds = Q\)
Differential form: ∇ . E = ρ/ε0
∇ . D = ρ
Gauss law for magnetic field:
Gauss's law for magnetism is one of the four Maxwell's equations. It states that the magnetic field B has divergence equal to zero, in other words, that it is a solenoidal vector field.
It is equivalent to the statement that magnetic monopoles do not exist.
Integral form: \(\smallint B.nds = 0\)
\(\smallint H.nds = 0\)
Differential form: ∇ . H = 0When the separation between two charges is made four times, the force between them
Coulomb’s law:
The force between two point charges is along the line joining between them is directly proportional to the product of point charges and inversely proportional to the square of the distance between them:
\(F\propto \frac{{{Q}_{1}}{{Q}_{2}}}{{{R}^{2}}}\)
Q1 & Q2 must be static or at rest.
The distance (R) between two-point charges R is very large compared to the dimension of a charged body
\(∴ F=\frac{K{{Q}_{1}}{{Q}_{2}}}{{{R}^{2}}}\)
Where \(K=\frac{1}{4\pi {{\varepsilon }_{o}}}=9\times {{10}^{9}}m/F\)
\(F=\frac{1}{4\pi {{\varepsilon }_{o}}}\frac{{{Q}_{1}}{{Q}_{2}}}{{{R}^{2}}}\)
\({{\varepsilon }_{o}}=\frac{1}{4\pi f}\frac{{{Q}_{1}}{{Q}_{2}}}{{{R}^{2}}}\frac{{{C}^{2}}}{N-{{m}^{2}}}\)
\({{\varepsilon }_{o}}=8.854\times {{10}^{-12}}\cdot \frac{F}{m}=\frac{{{10}^{-9}}}{36\pi }\frac{F}{m}\)
Coulomb force follows the superposition principle
Application:
When the separation (R) between two charges is made four times, then force between them is
\(F\propto \frac{{{}_{1}}}{{{R}^{2}}}\)
∴ Force is decreases sixteen times
Electromotive force, potential difference, and electrical potential are different forms of electric voltage. Therefore, the SI unit of all these is Volts (V)
Electric flux has an SI unit of volt-meters (V-m)
So, electric flux is different from others.Which one of the following is not a characteristic of magnetic flux?
Concept :
Magnetic flux is given by: Φ_{B} = BA cos θ where
Magnetic flux is usually measured with a flux meter. The SI and CGS unit of magnetic flux is given below:
Magnetic Flux Density (B) is defined as the force acting per unit current per unit length on a wire placed at right angles to the magnetic field.
Properties Of Magnetic Flux
Important Terms :
Air Gap: The distance between the north and south poles of a magnetic circuit. In conducting pull tests this is the distance between the working surface of the magnet and the testing apparatus.
Coercive Force, Hc: The intensity of a magnetic field required to reduce to zero the residual magnetism of a substance.
Curie Temperature: The temperature that a magnetic substance loses its magnetic properties.
Demagnetizing Force: A magnetized force applied in a direction that reduces the field in a magnetized material.
Gauss: The unit of magnetic induction or magnetic flux density used to measure magnetic field strength.
Gaussmeter: An instrument used to measure the intensity of a magnetic field.
Gradient: Indicates the change in magnetic strength between points measured at different distances perpendicular to the magnetic field.
Isotropic: (non-oriented) A material with no preferred direction of orientation resulting in the same magnetic characteristics through any axis.
Magnet: A material that has the property, either natural or induced, of attracting iron or steel.
Magnetic Field: The space around a magnet in which the magnetic force can be detected.
Magnetic Flux: The total magnetic induction across or through a specified area.
Magnetic Induction, B: The production of magnetic properties in a magnetizable substance when placed in a magnetic field.
Magnetic Lines Of Force: A series of invisible lines passing from one pole to another of a magnet, which taken together form the magnetic field.
Magnetic Saturation: The maximum amount of magnetic energy that can be absorbed by a magnetic substance.
Oersted: The unit of magnetic intensity in the cgs ( centimeter-gram-second ) system that
describes magnetic force.
Reach Out: The distance in which a magnetic field will extend from the magnet source.
Residual Magnetism: Small amounts of magnetism that remain in a material after being exposed to magnetic force.
Concept:
Force (F) ∝ q1 × q2
\(F \propto \frac{1}{{{r^2}}}\)
\(F = K\frac{{{q_1} \times {q_2}}}{{{r^2}}}\)
Where K is a constant = 9 × 109 Nm2/C2
Explanation:
Case 1:
Given that charges are: Q_{1} and Q_{2}
Distance = d
Force (F_{1}) = 20 mN
\(F \propto \frac{{{Q_1}{Q_2}}}{{{d^2}}}\)
Case 2:
The charges of both particles are doubled and distance between them is also doubled.
Given that charges are: 2Q_{1} and 2Q_{2}
Distance = 2d
\(F \propto \frac{{2{Q_1} \times 2{Q_2}}}{{{{\left( {2d} \right)}^2}}} = \frac{{{Q_1}{Q_2}}}{{{d^2}}}\)
Therefore, the force remains same and it is equal to 20 mN
Coulomb’s law:
When two charged particles of charges q1 and q2 are separated by a distance r from each other then the electrostatic force between them is directly proportional to the multiplication of charges of two particles and inversely proportional to the square of the distance between them.
Force (F) ∝ q1 × q2
\(F \propto \frac{1}{{{r^2}}}\)
\(F = K\frac{{{q_1} \times {q_2}}}{{{r^2}}}\)
Where K is a constant = 9 × 109 Nm2/C2
Calculation:
The given situation can be visualized as shown:
The force on Q because of q_{1} will be:
\({F_{Q{q_1}}} = \frac{{kQ{q_1}}}{{{a^2}}}\)
Similarly, the force on Q because of q_{2} will be:
\({F_{Q{q_2}}} = \frac{{kQ{q_2}}}{{{{\left( {3a} \right)}^2}}}\)
For the net charge at Q to be zero, we write:
\(\frac{{kQ{q_1}}}{{{a^2}}} + \frac{{kQ{q_2}}}{{9{a^2}}} = 0\)
\(\frac{{{q_1}}}{{{a^2}}} = \frac{{ - {q_2}}}{{9{a^2}}}\)
q_{2} = -9q_{1}