Answered: ©theexpertta.ce lill

Science

Physics

©theexpertta.ce lill

University Physics Volume 1

18th Edition

ISBN:9781938168277

Author:William Moebs, Samuel J. Ling, Jeff Sanny

Publisher:William Moebs, Samuel J. Ling, Jeff Sanny

Chapter1: Units And Measurement

Section: Chapter Questions

Problem 88CP: The first atomic bomb was detonated on July 16, 1945, at the Trinity test site about 200 mi south of...

See similar textbooks

Related questions

Q: A 41.0 mA current is carried by a uniformly wound air-core solenoid with 445 turns, a 12.0 mm…

Q: The tightly wound toroidal solenoid is one of the few configurations for which it is easy to…

A: The tightly wound toroidal solenoid field lines are circular along the length of the toroid. And…

Q: An air-core solenoid contains 1060 m of wire per meter and has a cross-sectional area of 0.23 m. A…

A: Given: The number of turns per length in the solenoid: n = 1060 m-1. The cross-sectional area of…

Q: A small thin coil with N2 loops, each of area A2, is placed inside a long solenoid, near its center.…

A: Mutual inductance of two coils linked with each other is given by:- M= (N2ØB_2)/i1 Where; ØB_2= flux…

Q: A solenoid passes through the center of a wire loop, as shown in (Figure 1). The solenoid has 1200…

Q: 2) A square loop of side 2 m is located in the plane x = 0 as shown in figure. A nonuniform magnetic…

Q: A long solenoid has a square cross-section with sides of length l, n turns per length has the z-axis…

Q: A long straight wire and a small rectangular wire loop lie in the same plane, as in the figure.…

Q: The inductor in (Figure 1) is a 9.0-cm-long, 2.0-cm-diameter solenoid wrapped with 300 turns.

A: Given: l = 9cm = 0.09m D = 2cm , R = 1cm = 0.01m N = 300

Q: Using these data, discuss this situation in your group and determine the emf of the battery, the…

Q: An inductor is built using a circular coil that is l 0.50 m long, has radius R 10.0 cm, and consists…

A: a)Self inductance of the coil is,

Q: 6. Consider a solenoid with length l, cross sectional area A, and N windings. (a) What is the…

A: Resistance= 5 ohms N= 2000 turn r= 5mm L= 4 cm

Q: Suppose a 57.5 cm long, 11.5 cm diameter solenoid has 1000 loops. Part (a) Calculate the…

A: Answer:- Given The length of the solenoid(xo) = 57.5 cm = 0.575 m The diameter of the…

Q: A 40-turn, 4.0-cm-diameter coil with R= 0.402 surrounds a 3.0-cm-diameter solenoid. The solenoid is…

Q: Q3: A long solenoid, of radius Scm, is driven by an altemating current, so that radius - cm and…

Q: Two solenoids A and B, spaced close to each other and sharing the same cylindrical axis, solenoid A…

A: Given:

Q: The long solenoid S shown (in cross section) in Figure 9 has 220 turns/cm and carries a current I =…

A: N=Number of turns=130n=Number of turns per unit length=220 turns/cmd=Diameter of coil=2.1…

Q: (a) Find the inductance of the solenoid. (b) If the solenoid has 350 turns, what is the magnetic…

A: a) The solenoid’s self-induced emf can be represented as, Here, L, I, and t represent the…

Q: A 298 turn solenoid has a radilus of 4.95 cm and a length of 19.5 cm. (a) Find the inductance of the…

A: Formula of inductance of solenoid is as follows: L=μ0N2Al…

Q: A 510-turn solenoid has a radius of 7.95 mm and an overall length of 13.2 cm. (a) What is its…

A: Given : Number of turns, N = 510 radius, r = 7.95 mm Length, l = 13.2 cm = 132 mm Area , A = πr2…

Q: Chapter 30, Problem 076 A coil C of N = 121 turns is placed around a long solenoid S of radius R =…

Q: Experimentally, the magnetic flux ΦB in a solenoid was measured for various currents ?. The data was…

A: Given data The current at the first point is given as I1 = 0.750 A. The current at the second point…

Q: Edhina An aluminum ring of radius n = 5.00 cm and a resistance of 2.75 x 10 N is placed around one…

Q: Find the inductance and induced emf for a solenoid with a cross-sectional area of 1.3 × 10-3m2. The…

A: Given: Cross-sectional area of solenoid = 1.3×10-3 m2. Length of solenoid l = 9.0 cm. = 9×10-2 m.…

Q: A solenoid of length 6.9x10-2 m and cross-sectional area of 6.8x10-5 m2 contains 6350 turns per…

A: Given that Length of the solenoid l = 6.9 × 10-2 m Cross sectional area of the solenoid A = 6.8 ×…

Q: A 497-turn solenoid has a radius of 9.00 mm and an overall length of 13.2 cm. (a) What is its…

Q: A long fund rival solenoid with 100 turns per cm has a radius of 1.6 cm. Assume the magnetic field…

Q: A air-filed solenoid 18.3 cm diameter and length 23 cm, has 120 turns. If the current increases from…

A: Diameter , d = 18.3 cm = 0.183 m Length , L = 23 cm = 0.23 m Number of turns , N = 120 turns…

Q: Tarik winds a small paper tube uniformly with 173 turns of thin wire to form a solenoid. The tube's…

A: Given data : Number of turns, N = 173 Diameter of the tube, D = 7.95 mm =…

Q: a solenoid with a cross-sectional of 5cm^2 has 1000 turns and is 35 cm long. a second coil is…

A: (a) Find the mutual inductance of the inductor. M=μ0N1N2Al=4π×10-7 H/m10004005×10-4 m20.35…

Q: A coil of 40 turns is wrapped around a long solenoid of cross-sectional area 7.5 × 10−3 m2. The…

A: The relation between magnetic flux (Φ21) and mutual inductance (M) is given by- M=N2Φ21I1Here,N2 is…

Concept explainers

Voltage Across Inductor

An actor is a passive electrical component that consists of a coil bent in the form of wire which uses electromagnetism to produce electric current through the coil. When the electric current flows through a conductor that is the coil, a magnetic flux is developed around the conductor. This flow of current through the coil creates a magnetic field based on Fleming's Right-hand thumb rule. If there is any decrease in the current passing through the inductor, then the magnetic field will also decrease and energy is released through the generation of electric current. There is also a secondary voltage inducing in the same coil due to the magnetic flux, since it opposes any changes in the electric current flowing through the circuit. An inductor is also called choke, coil, or solenoid. A choke is designed to block certain frequencies when the DC current passes. A winding coil is coated with insulation to give extra insulation and protect the circuit. They are usually tightly wrapped around a central core which is cylindrical in nature so that the magnetic flux can be induced. The solenoid is a three-dimension coil that is used to exhibit electromagnetism when the magnetic field is developed due to the current flowing through the coil. The sign convention of the inductor is the same as the power dissipating device. When the current flows into the positive side of the voltage across the inductor, the value is positive and the inductor dissipates power whereas when the inductor releases the power back to the circuit, the current has a negative sign convention. 

Question

100%

An air-core solenoid has N = 1525 turns, d = 0.37 m length, and cross sectional area A = 0.066 m². The current flowing through the solenoid is I = 0.764 A.

Express the inductance of the solenoid, L, of the solenoid in terms of N, d, and A.

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

This is a popular solution!

SEE SOLUTION Check out a sample Q&A here

Step 1

VIEW

Step 2

VIEW

Trending now

This is a popular solution!

Step by step

Solved in 2 steps with 2 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.

Similar questions

Hello, I need help with this question. Pls & thank you!
asnwer d
MM12.. no handwritten...
Please give me both answers otherwise leave it... Remember give me both answers.. Thank you
need answer asap...if you can answer correctly..ill upvite you
plssss solve correctly
please answer all aprts of quesrtions
12) Please need the correct to this problem with full explanation and methods how you get the answer, please.
Plz thorough explain...
The first atomic bomb was detonated on July 16, 1945, at the Trinity test site about 200 mi south of Los Alamos. In 1947, the U.S. government declassified a film reel of the explosion. From this film reel, British physicist G.I. Taylor was able to determine the rate at which the radius of the fireball from the blast grew. Using dimensional analysis, he was then able to deduce the amount of energy released in the explosion, which was a closely guarded secret at the time. Because of this, Taylor did not publish his results until 1950. This problem challenges you to recreate this famous calculation. (a) Using keen physical insight developed from years of experience, Taylor decided the radius rof the fireball should depend only on time since the explosion, t, the density of the air, , and the energy of the initial explosion, E. Thus, he made the educated guess that r=kEabtcfor some dimensionless constant kand some unknown exponents a,b, and c. Given that [E]=ML2T-2 , determine the values of the exponents necessary to make this equation dimensionally consistent. (Hint: Notice the equation implies that k=rEabtcand that [k]=1 ). (b) By analyzing data from high-energy conventional explosives, Taylor found the formula he derived seemed to be valid as long as the constant khad the value 1.03. From the film reel, he was able to determine many values of rand the corresponding values of t. For example, he found that after 25.0 ms, the fireball had a radius of 130.0 m. Use these values, along with an average air density of 1.25kg/m3 , to calculate the initial energy release of the Trinity detonation in joules (J). (Hint: To get energy in joules, you need to make sure all the numbers you substitute in are expressed in terms of SI base units.) (c) The energy released in large explosions is often cited in units of “tons of TNT” (abbreviated “t TNT”), where 1 t TNT is about 4.2 GJ. Convert yow answer to (b) into kilotons of TNT (that is, kt TNT). Compare your answer with the quick-and-duty estimate of 10 kt TNT made by physicist Enrico Fermi shortly after witnessing the explosion from what was thought to be a safe distance. (Reportedly, Fermi made his estimate by dropping some shredded bits of paper right before the remnants of the shock wave hit him and looked to see how far they were carried by it.)
Pls explain in a few sentences.
Neeed help...