©theexpertta.ce lill

Q: In the circuit shown in (Figure 1), E = 56.0 V, R₁ = 36.0 , R₂ = 22.0 , and L= 0.290 H. Part A…

A: Givenε=56 VR1=36 ΩR2=22 ΩL=0.290 H

Q: lowing through a solenoid of inductance 7.4 H.    Part(a): Determine the magnetic flux through the…

A: The Current in the solenoid of inductance is I = 3.9 A  The Inductance L = 7.4 H  Find:- Part(a):…

Q: A solenoid with 485 turns has a length of 7.00 cm and a cross-sectional area of 3.50 x 10-9 m2. Find…

A: a) The inductance of the solenoid can be obtained as L=μ∘N2AlSubstituting the given values :L=(4π x…

Q: A solenoid with 455 turns has a length of 8.00 cm and a cross-sectional area of 2.90 ✕ 10−9 m2. Find…

A:

Q: A 4.10 x 10 -5 H solenoid is constructed by wrapping 77 turns of wire around a cylinder with a…

A:

Q: At time t = 0, a 12.0 V potential difference is suddenly applied to the leads of a coil of…

A:

Q: the solenoid's inductance (in H)   H (b) the average emf around the solenoid (in V)

A:

Q: There is a tank of width and length of w and a height of 0.6 m. This tank is wrapped around with…

A:

Q: A solenoid with 435 turns has a length of 7.00 cm and a cross-sectional area of 3.10 x 10-9 m². Find…

A:

Q: Self-inductance (L) of a 40.0 cm long solenoid containing N = 400 turns, and cross- sectional area…

A: Solution:-Given thatLength of solenoid (l)=40 cm=0.4 mNumber of turns (N)=400Cross sectional area…

Q: A solenoid with 485 turns has a length of 5.00 cm and a cross-sectional area of 2.90 ✕ 10−9 m2. Find…

A:

Q: A solenoid with 435 turns has a length of 5.50 cm and a cross-sectional area of 2.60 ✕ 10−9 m2. Find…

A: Given value--- number of turns = 435. Length = 5.5 cm. Area = 2.6 × 10-9 m2 . change in current = 4…

Q: N2 EN1

A: Given: The permeability of the magnetic core is  µ= ∞ The area of the cross-section of the core is…

Q: A long solenoid (n=2312 turns/m) has a cross-sectional area of 0.55 m² and a current given by I(t) =…

A: Given that Number of turns of long solenoid,n=2312 turns/m Cross-sectional area of long…

Q: A solenoid has a winding of 400 turns, has a length of 30.0 cm and an area of 5.00 cm2. Find the…

A: Step 1: Step 2: Step 3: Step 4:

Q: A 5.70 x 10 -5 H solenoid is constructed by wrapping 80 turns of wire around a cylinder with a…

A:

Q: An air-core solenoid has N = 195 turns, d = 0.505 m length, and cross sectional area A = 0.034 m2.…

A: The number of turns in the solenoid: The length of the solenoid: The cross-section area: The current…

Q: Consider the RL circuit shown in the figure, where R = 16.3 0. 03 The switch S is closed at t = 0.…

A: Given that,In a RL circuitR=16.3 Ωtime t=3.00 msLet, final value of current is I0Thus, according to…

Q: goes L n

A: Initial value of current,                                         I1 = 0.200 A  Final value of…

Q: An air-core solenoid has N = 575 turns, d = 0.37 m length, and cross sectional area A = 0.074 m2.…

A:

Q: A solenoid with 425 turns has a length of 5.50 cm and a cross-sectional area of 2.60 ✕ 10−9 m2. Find…

A: Given quantities: Number of turns (N) = 425 Length of the solenoid (d) = 5.50 cm Area of…

Q: A solenoid with 495 turns has a length of 6.50 cm and a cross-sectional area of 3.10 x 10-9 m². Find…

A: If the number of loops is increased then magnetic fields are also increased in the case of…

Q: A solenoid with 455 turns has a length of 6.50 cm and a cross-sectional area of 3.50 ✕ 10−9 m2. Find…

A: GivennN=455 turnsl=6.50 cm=0.065 mA=3.50×10-9 m2Current change from +2.50 A to -2.50 ATime…

Q: A solenoid with 435 turns has a length of 5.00 cm and a cross-sectional area of 3.10 x 109 m². Find…

A: Given value--- number of turns = 425 . length = 5 cm. Area = 3.10 × 10-9 m2 . current changes from…

Q: Determine the inductance of a solenoid with 610 turns in a length of 29 cm. The circular cross…

A:

Q: A solenoid of radius 2.50 cm has 400 turns and a length of 20.0 cm. Find its inductance

A: We have given  Radius  r = 2.50 cm             r = 2.50 × 10-2m  Number of turns  N = 400  Length of…

Q: Current through an inductor(L= 0.1 H) is given by i = (t²-6t) A. Find the magnitude of the induced…

A:

Q: A solenoid with 495 turns has a length of 8.00 cm and a cross-sectional area of 3.20 ✕ 10−9 m2. Find…

A: Required :  (a)the solenoid's inductance (in H)H(b)the average emf around the solenoid (in V)V

Q: Two solenoids A and B, spaced close to each other and sharing the same cylindrical axis, have 380…

A:

Q: A small loop of area 6.8 mm2 is placed inside a long solenoid that has 854 turns/cm and carries a…

A: Magnetic field produced by solenoid,B=μnI      Equation 1Given, μ=4π×10-7n=854×1/10-2 mI=1.28

Q: A solenoid with 455 turns has a length of 6.50 cm and a cross-sectional area of 2.60 ✕ 10−9 m2. Find…

A:

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

A: Given:

Q: A solenoid with 485 turns has a length of 5.50 cm and a cross-sectional area of 3.10 ✕ 10−9 m2. Find…

A: The inductance of the solenoid can be obtained as

Q: A solenoid has an inductance L-2.60 H. The time-dependent induced emf in the solenoid is given by…

A: Given,           The inductance of solenoid is L=2.60 H.          The time dependent induced emf as…

Q: A solenoid of radius 1.50 cm has 600 turns and a length of 25.0 cm. (a) Find its inductance. mH (b)…

A: To calculate the inductance of a solenoid, we can use the formula for the inductance of a solenoid:

Q: with 435 turns has a length of 6.50 cm and a cross-sectional area of 3.10 ✕ 10−9 m2. Find the…

A: Given values:- Number of turns,(N)=435 Turns ----(1) Length of solenoid,(l)=6.50 cm=6.50 x 10-2 m…

Q: self-inductance of the solenoid

A:

Q: solenoid with 525 turns has a length of 5.50 cm and a cross-sectional area of 3.00 ✕ 10−9 m2. Find…

A: number of turns = N = 525 length = l = 5.5 cm area = A = 3 X 10-9 m2

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: How many turns should a solenoid of cross-sectional area 3.9×10−2 m2 and length 0.31 m have if its…

A:

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.