Which condition is required for Coulomb's law to hold true?

1 answer:

7 0

The correct answer is:
Point charges must be in a vacuum.

In fact, the usual form for of the Coulomb's law is:
 $F= \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2}$
where
 $\epsilon_0$ is the permittivity of free space
q1 and q2 are the two charges
q is the separation between the two charges

However, this formula is valid only if the charges are in vacuum. If they are in a material medium, the law is modified as follows:
 $F= \frac{1}{4 \pi \epsilon_0 \epsilon_r} \frac{q_1 q_2}{r^2}$
where $\epsilon_r$ is the relative permittivity, which takes into account the dielectric effects of the material.

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The effects due to the interaction of a current-carrying loop with a magnetic field have many applications, some as common as th

antoniya [11.8K]

Answer:

a) τ = 1.039*10⁻⁴N-m

b) The net torque acting on the loop is zero, but the loop continues to rotate in a counterclockwise direction.

Explanation:

A) Given

I = 0.5 A

B = 0.3 T

a = 4 cm = 0.04 m

b = 2 cm = 0.02 m

θ = 30°

The torque τ acting on a current-carrying loop of area A due to the interaction of the current I flowing through the loop with a magnetic field of magnitude B is given by

τ = I*B*A*Sin∅

where ∅ is the angle between the normal to the loop and the direction of the magnetic field.

The area A of a rectangular loop of wire with height 4.00 cm and horizontal sides 2.00 cm can be obtained as follows

Aloop = a*b ⇒ Aloop = 0.04 m*0.02 m = 8*10⁻⁴m²

Recalling that θ is the angle between the sides of length b and B and if we consider the normal to the loop, the angle ∅ between the normal and the magnetic field is given by

∅ = 90°-θ ⇒ ∅ = 90°-30° = 60°

Then, the torque will be

τ = (0.5 A)*(0.3 T)*(8*10⁻⁴m²)*Sin60° = 1.039*10⁻⁴N-m

b) We have to get the net torque τ about the vertical axis of the current loop due to the interaction of the current with the magnetic field.

The angle ∅ between the normal to the loop and the magnetic field when the horizontal sides of the loop of length b are perpendicular to B is

∅ = 0°

Then

τ = (0.5 A)*(0.3 T)*(8*10⁻⁴m²)*Sin 0° = 0 N-m

We can say that the net torque acting on the loop is zero, but the loop continues to rotate in a counterclockwise direction.

5 0

4 years ago

A pitcher can throw a baseball at about 90 miles/hour (about 39.6 m/s). What is the ratio of the kinetic energy to the rest ener

tester [92]

Answer:

$8.71*10^{-15}$

Explanation:

The rest energy is the energy associated with the base ball at zero velocity which is expressed as

$Rest energy=mc^{2}\\$

Note: the speed C is the speed of light which is expressed as 3*10^8

while the kinetic energy is the energy associated with the ball during its motion and is expressed as

$Kinetic energy=1/2mv^{2}$

the ratio can be expressed as

$\frac{Kinetic Energy}{Rest Energy}=\frac{\frac{1}{2}mv^{2} }{mc^{2}}\\since V=39.6m/s\\C=3*10^{8}m/s\\\frac{Kinetic Energy}{Rest Energy}=\frac{\frac{1}{2}39.6^{2} }{(3*10^8)^{2}} \\\frac{Kinetic Energy}{Rest Energy}=\frac{784.08}{9*10^{16}} \\\frac{Kinetic Energy}{Rest Energy}=8.71*10^{-15}$