The Maxwell equation derived from Gauss's Law in magnetism is a)oint vec(B) vec(ds)=mu_0 I+ epsilon_0 mu_0 (d Phi_E)/(dt) b)oint vec(B) vec(dA)=0 c)oint vec(E) vec(dA)=(Q_(in))/(epsilon_0) d)oint vec(E) vec(ds)=-(d Phi_m)/(dt)

Lilliana Livingston 2022-07-20 Answered
The Maxwell equation derived from Gauss's Law in magnetism is
a) B d s = μ 0 I + ϵ 0 μ 0 d Φ E d t
b) B d A = 0
c) E d A = Q i n ϵ 0
d) E d s = d Φ m d t
You can still ask an expert for help

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Answers (1)

Brendon Bentley
Answered 2022-07-21 Author has 11 answers
According to Gauss`s law for magnetism the net magnetic flux through any closed surface is zero
So, Φ = d ϕ = B d A = 0
where, ϕ=magnetic flux
B=magnetic field
dA= area element
Hence option b is correct
option (a) is incorrect because it is ampere maxwell law
option (c) is incorrect because it is Gauss`s law for electricity and option (d) is also incorrect because it is faraday`s law
Not exactly what you’re looking for?
Ask My Question

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

You might be interested in

asked 2021-02-06

Starting with an initial speed of 5.00 m/s at a height of 0.300 m, a 1.50 kg ball swings downward and strikes a 4.60kg ballthat is at rest, as the drawing shows. a. using the principle of conservation of mechanicalenergy,find the speed of the 1.50 kg ball just before impact b. assuming that the collision is elastic, find the velocities( magnitude and direction ) of both balls just after thecollision c. how high does each abll swing after the collision, ignoringair resistance?

asked 2021-01-13

A radar station, located at the origin of xz plane, as shown in the figure , detects an airplane coming straight at the station from the east. At first observation (point A), the position of the airplane relative to the origin is RA. The position vector RA has a magnitude of 360m and is located at exactly 40 above the horizon. The airplane is tracked for another 123} in the vertical east-west plane for 5.0s, until it has passed directly over the station and reached point B. The position of point B relative to the origin is RB (the magnitude of RB is 880m). The contact points are shown in the diagram, where the x axis represents the ground and the positive z direction is upward.
image
Define the displacement of the airplane while the radar was tracking it: RBA=RBRA. What are the components of RBA
Express RBA in meters as an ordered pair, separating the x and z components with a comma, to two significant figures.

asked 2021-05-04

Until he was in his seventies, Henri LaMothe excited audiences by belly-flopping from a height of 13 m into 35 cm. of water. Assuming that he stops just as he reaches the bottom of the water and estimating his mass to be 72 kg, find the magnitudes of the impulse on him from the water.

asked 2021-03-05

A cylindrical capacitor has an inner conductor of radius 1.5 mm andan outer conductor of radius 3.4 mm.The two conductors are separated by vacuum, and the entire capacitor is 3.0 m long. 
a) What is the capacitance per unit length? 
b)The potential of the inner conductor is 350 mV higher than that of the outer conductor. Find the charge (magnitude and sign) on both conductors. Find the charge (magnitude and sign) on both conductors.

asked 2022-05-19
Why are there more diamagnetic substances in nature than there are paramagnetic or ferromagnetic substances?
asked 2022-05-18
The inverse square law for an electric field is:
E = Q 4 π ε 0 r 2
Here:
Q ε 0
is the source strength of the charge. It is the point charge divided by the vacuum permittivity or electric constant, I would like very much to know what is meant by source strength as I can't find it anywhere on the internet. Coming to the point an electric field is also described as:
E d = F d Q = Δ V
This would mean that an electric field can act only over a certain distance. But according to the Inverse Square Law, the denominator is the surface area of a sphere and we can extend this radius to infinity and still have a value for the electric field. Does this mean that any electric field extends to infinity but its intensity diminishes with increasing length? If that is so, then an electric field is capable of applying infinite energy on any charged particle since from the above mentioned equation, if the distance over which the electric field acts is infinite, then the work done on any charged particle by the field is infinite, therefore the energy supplied by an electric field is infinite. This clashes directly with energy-mass conservation laws. Maybe I don't understand this concept properly, I was hoping someone would help me understand this better.
asked 2022-01-11

Along coaxial cable consists of an inner cylindrical conductor with radius a and an outer coaxial cylinder with inner radius b and outer radius c. The outer cylinder is mounted on insulating supports and has no net charge. The inner cylinder has a uniform positive charge per unit length λ. Calculate the electric field 
a) at any point between the cylinders a distance r from the axis and 
b) at any point outside the outer cylinder. 
c) Graph the magnitude of the electric field as a function of the distance r from the axis of the cable, from r= 0 to r= 2c. 
d) Find the charge per unit length on the inner surface and on the outer surface of the outer cylinder.

New questions