PHYSICS COLLEGE

A long, thin superconducting wire carrying a 17 A current passes through the center of a thin, 3.0-cm-diameter ring. A uniform electric field of increasing strength also passes through the ring, parallel to the wire. The magnetic field through the ring is zero.a. At what rate is the electric field strength increasing?
b. is the electric field in the direction of the current or opposite to the current?

Answers

Answer 1

Answer:

Answer:

$(dE)/(dt) =- 2.72 *10^(15) \ N/C \cdot s$

The direction of the electric field is opposite that of the current

Explanation:

From the question we are told that

The current is $I = 17\ A$

The diameter of the ring is $d = 3.0 \ cm = 0.03 \ m$

Generally the radius is mathematically represented as

$r = (d)/(2)$

$r = (0.03)/(2)$

$r = 0.015 \ m$

The cross-sectional area is mathematically represented as

$A = \pi r^2$

=> $A = 3.142 * (0.015^2)$

=> $A = 7.07 *10^(-4 ) \ m^ 2$

Generally according to ampere -Maxwell equation we have that

$\oint \= B \cdot \= ds = \mu_o I + \epsilon_o \mu _o( d \phi )/(dt )$

Now given that $\= B = 0$ it implies that

$\oint \= B \cdot \= ds = 0$

$\mu_o I + \epsilon_o \mu _o( d \phi )/(dt ) = 0$

Where $\epsilon _o$ is the permittivity of free space with value $\epsilon_o = 8.85*10^(-12 ) \ m^(-3) \cdot kg^(-1)\cdot s^4 \cdot A^2$

$\mu_o$ is the permeability of free space with value

$\mu_o = 4\pi * 10^(-7) N/A^2$

$\phi$ is magnetic flux which is mathematically represented as

$\phi = E * A$

Where E is the electric field strength

$\mu_o I + \epsilon_o \mu _o ( d [EA] )/(dt ) = 0$

=> $(dE)/(dt) =- (I)/(\epsilon_o * A )$

=> $(dE)/(dt) =- (17)/(8.85*10^(-12) * 7.07*10^(-4) )$

=> $(dE)/(dt) =- 2.72 *10^(15) \ N/C \cdot s$

The negative sign shows that the direction of the electric field is opposite that of the current

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The position of a particle is given by the function x=(5t3−8t2+12)m, where t is in s. at what time does the particle reach its minimum velocity?

Answers

The particle reach its minimum velocity at time 1.06 sec.

The function is given as

x=5t^3-8t^2+12

Differentiating the above equation with respect to time, to obtain the velocity

dx/dt=v=15t^2-16t

For maximum and minimum values, put dx/dt=0

15t^2-16t=0

On solving the equation, t=0, 1.06

Therefore at time t=1.06 sec, the particle has the minimum value of velocity.

The particle reaches its minimum velocity at t = 0 s or t = 16/15 s

Further explanation

Acceleration is rate of change of velocity.

$\large {\boxed {a = (v - u)/(t) } }$

$\large {\boxed {d = (v + u)/(2)~t } }$

a = acceleration ( m/s² )

v = final velocity ( m/s )

u = initial velocity ( m/s )

t = time taken ( s )

d = distance ( m )

Let us now tackle the problem!

Given:

$x = ( 5t^3 - 8t^2 + 12) ~ m$

To find the velocity function, we will derive the position function above.

$v = (dx)/(dt)$

$v = 5(3)t^(3-1) - 8(2)t^(2-1)$

$v = ( 15t^2 - 16t ) ~ m/s$

Next to calculate the time to reach its minimum speed, then v = 0 m/s

$0 = ( 15t^2 - 16t )$

$0 = t( 15t - 16)$

$\large {\boxed {t = 0 ~s ~ or ~ t = 16/15 ~ s} }$

Learn more

Velocity of Runner : brainly.com/question/3813437
Kinetic Energy : brainly.com/question/692781
Acceleration : brainly.com/question/2283922
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Answer details

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle

A student and his lab partner create a single slit by carefully aligning two razor blades to a separation of 0.530 mm. When a helium–neon laser at 543 nm illuminates the slit, a diffraction pattern is observed on a screen 1.55 m beyond the slit. Calculate the angle θdark to the first minimum in the diffraction pattern and the width of the central maximum.

Answers

Answer:

angle = 0.058699 degree

width of central maximum is 3.170566 × $10^(-3) ) mExplanation:Given data separation d = 0.530 mm = 0.530×[tex]10^(-3)$ m

distance D = 1.55 m

wavelength w = 543 nm = 543× $10^(-9)$ m

to find out

angle θ and width of the central maximum

solution

we know according to first condition first dark that mean

wavelength = dsinθ

so put value and find θ

543× $10^(-9)$ = 0.530× $10^(-3)$ ×sinθ

sinθ = 543× $10^(-9)$ / 0.530× $10^(-3)$

sinθ = 1.02452 × [tex]10^{-3}

θ = 0.058699 degree

and

we can say

tanθ = y/D

here y is width of central maximum Y = 2y

put all value we get y

so y = D tanθ

y = 1.55 (tan0.0586)

y = 1.58528 × [tex]10^{-3} m =

so Y = 2 ( 1.58528 × [tex]10^{-3} )

so width of central maximum is 3.170566 × [tex]10^{-3} ) m

Which volcanoes are formed by pyroclastic deposits? Select all that apply. A. cinder cone B. stratovolcano C. shield volcano

Answers

Answer:

Its a cinder cone cause after it all falls down to make deposits.

cinder cone and stratovolcano

the period of the satellite is exact 42.391 hours, the earth's mass is 5.98 kg and the radius of th earth is 3958.8 miles, what is the distance of the satellite from the surface of the earth in miles?

Answers

Answer:

As the mass is not written well, i will use the equation in terms of the gravitational acceleration:

The equation for the period of a satellite is:

$T = 2*pi*\sqrt{(r^3)/(g*R^2) }$

We want to find r, so isolating r we get:

$r = \sqrt[3]{x ((T)/(2*pi) )^2*g*R^2}$

Where:

T = period.

r = radius of the satellite.

R = radius of the planet.

g = gravitational acceleration of the planet.

pi = 3.14159...

g = 78999.64 mi/h^2 (value of a table)

T = 42.391 h.

R = 3958.8 miles

We can replace those values in the equation and get:

$r = \sqrt[3]{ ((42.391)/(2*3.14159) )^2*78999.64*(3958.8)^2} = 38,339.5 mi$

Now this value is measured from the center of the Earth, then the altitude of the satellite measured from the surface of the Earth will be:

H = r - R = 38,339.1mi - 3958.8mi = 34,380.3 mi

Two long parallel wires are a center-to-center distance of 1.30 cm apart and carry equal anti-parallel currents of 2.40 A. Find the magnitude of the magnetic field at the point P which is equidistant from the wires. (R = 5.00 cm).

Answers

Image is missing, so i have attached it

Answer:

19.04 × 10⁻⁴ T in the +x direction

Explanation:

We are told that the point P which is equidistant from the wires. (R = 5.00 cm). Thus distance from each wire to O is R.

Hence, the magnetic field at P from each wire would be; B = μ₀I/(2πR)

We are given;

I = 2.4 A

R = 5 cm = 0.05 m

μ₀ is a constant = 4π × 10⁻⁷ H/m

B = (4π × 10⁻⁷ × 2.4)/(2π × 0.05)

B = 9.6 × 10⁻⁴ T

To get the direction of the field from each wire, we will use Flemings right hand rule.

From the diagram attached:

We can say the field at P from the top wire will point up/right

Also, the field at P from the bottom wire will point down/right

Thus, by symmetry, the y components will cancel out leaving the two equal x components to act to the right.

If the mid-point between the wires is M, the the angle this mid point line to P makes with either A or B should be same since P is equidistant from both wires.

Let the angle be θ

Thus;

sin(θ) = (1.3/2)/5

θ = sin⁻¹(0.13) = 7.47⁰

The x component of each field would be:

9.6 × 10⁻⁴cos(7.47) = 9.52 × 10⁻⁴ T

Thus, total field = 2 × 9.52 × 10⁻⁴ = 19.04 × 10⁻⁴ T in the +x direction

Final answer:

The magnetic field at point P, which is equidistant from two long parallel wires with equal anti-parallel currents, is calculated using Ampere's law. The net magnetic field is zero because the fields due to each wire cancel each other at that point.

Explanation:

The question concerns the calculation of the magnetic field at a point equidistant from two long parallel wires that carry equal anti-parallel currents. According to the right-hand rule and Ampere's law, each wire generates a magnetic field that circles the wire. For two wires carrying currents in opposite directions, the magnetic fields at the midpoint between the wires will point in opposite directions, thus they will subtract from each other when calculating the total magnetic field at point P.

To find the magnetic field at point P, we use the formula for the magnetic field due to a long straight current-carrying wire: B = (μ₀I)/(2πd), where B is the magnetic field, μ₀ is the permeability of free space (4π x 10-7 T·m/A), I is the current, and d is the distance to the point of interest from the wire. In this case, the distance d will be the radius R = 5.00 cm since point P is equidistant from both wires.

Substituting the values into the formula, the magnetic field due to each wire at point P can be calculated. However, since the currents are anti-parallel, the net magnetic field at P would be the difference between the two fields, which is zero.

How much electrical energy is used by a 400 W toaster that is operating for 5minutes?
A. 2000 J
B. 75,000 J
C. 120,000 J
D. 300,000 J

Answers

The electrical energy used by a 400 W toaster that is operating for 5 minutes will be 120,000 J.Option C is correct.

What is the power output?

The rate of the work done is called the power output. It is denoted by P.Its unit of a watt. It is the ratio of the work done or the enrgy to the time period.

The given data in the problem is;

E is the electrical energy

P is the power output = 400 W

t is the time period = 5 minutes

The power output is given as;

$\rm P= (E)/(t) \n\n\ E= P * t \n\n\ E= 400 * 300 \n\n\ E=120,000 \ J$