Ch 30: Electromagnetic Induction

Knight Calc - Physics for Scientists and Engineers 5th Edition

Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook

Knight Calc 5th Edition

Ch 30: Electromagnetic Induction

Problem 69a

Chapter 30, Problem 69a

A 50 cm solenoid with 1000 turns has an inductance of 20 mH. What is the magnetic field strength inside the inductor when the current is 75 mA?

Verified step by step guidance

Step 1: Recall the formula for the inductance of a solenoid:

L = μ₀μr (N²A)/l

, where

L

is the inductance,

μ₀

is the permeability of free space,

μr

is the relative permeability,

N

is the number of turns,

A

is the cross-sectional area, and

l

is the length of the solenoid. However, for this problem, we are directly given the inductance and need to calculate the magnetic field strength.

Step 2: Use the formula for the magnetic field strength inside a solenoid:

B = μ₀μr (N/l)I

, where

B

is the magnetic field strength,

N

is the number of turns,

l

is the length of the solenoid, and

I

is the current.

Step 3: Convert the given values into SI units. The length of the solenoid is

l = 50 \, \(\text{cm}\) = 0.50 \, \(\text{m}\)

, the current is

I = 75 \, \(\text{mA}\) = 0.075 \, \(\text{A}\)

, and the number of turns is

N = 1000

. The permeability of free space is a constant:

μ₀ = 4π × 10^{-7} \, \(\text{T·m/A}\)

. Assume the relative permeability

μr

is 1 (for air or vacuum).

Step 4: Substitute the values into the formula for the magnetic field strength:

B = (4π × 10^{-7})(1000/0.50)(0.075)

. Simplify the expression step by step to calculate the magnetic field strength.

Step 5: Ensure the units are consistent and verify the calculation. The result will give the magnetic field strength

B

in teslas (T).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Inductance

Inductance is a property of an electrical component, typically a coil or solenoid, that quantifies its ability to store energy in a magnetic field when an electric current flows through it. It is measured in henries (H) and is defined as the ratio of the induced electromotive force (EMF) to the rate of change of current. In this case, the solenoid's inductance of 20 mH indicates its capacity to generate a magnetic field in response to the current.

Magnetic Field Strength

Magnetic field strength, often denoted as H, is a measure of the intensity of the magnetic field produced by an electric current. For a solenoid, it can be calculated using the formula H = nI, where n is the number of turns per unit length and I is the current. In this scenario, the magnetic field strength inside the solenoid can be determined by substituting the given values of current and the solenoid's dimensions.

Solenoid

A solenoid is a cylindrical coil of wire that generates a magnetic field when an electric current passes through it. The strength of the magnetic field inside a long solenoid is uniform and can be calculated using the formula B = μ₀nI, where B is the magnetic field, μ₀ is the permeability of free space, n is the number of turns per unit length, and I is the current. Understanding the properties of solenoids is essential for analyzing their behavior in electromagnetic applications.

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Related Practice

Textbook Question

BIO One possible concern with MRI (see Exercise 28) is turning the magnetic field on or off too quickly. Bodily fluids are conductors, and a changing magnetic field could cause electric currents to flow through the patient. Suppose a typical patient has a maximum cross-section area of 0.060 m². What is the smallest time interval in which a 5.0 T magnetic field can be turned on or off if the induced emf around the patient's body must be kept to less than 0.10 V?

Textbook Question

CALC FIGURE P30.67 shows the potential difference across a 20 mH inductor. The current through the inductor at t = 0 ms is 0.25 A. What is the current at t = 10 ms?

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Textbook Question

A 50 cm solenoid with 1000 turns has an inductance of 20 mH. Flipping a switch disconnects the inductor from the battery and connects it to a resistor. What is the value of the resistance if the magnetic field decreases by 50% in 150 μs?

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Textbook Question

CALC The current through inductance L is given by $I = I_{0} e^{- t / τ}$ . Find an expression for the potential difference ΔV_L across the inductor.

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Textbook Question

CALC The current through inductance L is given by $I = I_{0} e^{- t / τ}$ . Evaluate ΔV_L at t = 0, 1.0, and 3.0 ms if L = 20 mH, I₀ = 50 mA, and $τ$ = 1.0 ms.

Textbook Question

One way to measure the strength of a magnetic field is with a flip coil. Suppose a 200-turn, 4.0-cm-diameter coil with a resistance of 2.0 Ω is connected to a ballistic galvanometer, a device that measures the total charge passing through. The coil is held perpendicular to the field, then quickly flipped 180° so that the opposite side is facing the magnetic field. Afterward, the galvanometer reads 7.5 μC. What is the field strength? Hint: Use I = dq/dt to relate the net change of flux to the amount of charge that flows through the galvanometer.

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