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 6

Chapter 30, Problem 6

An equilateral triangle 8.0 cm on a side is in a 5.0 mT uniform magnetic field. The magnetic flux through the triangle is 6.0 μWb. What is the angle between the magnetic field and an axis perpendicular to the plane of the triangle?

Verified step by step guidance

Step 1: Recall the formula for magnetic flux, which is given by Φ = B × A × cos(θ), where Φ is the magnetic flux, B is the magnetic field strength, A is the area of the surface, and θ is the angle between the magnetic field and the normal (perpendicular) to the surface.

Step 2: Calculate the area of the equilateral triangle. The formula for the area of an equilateral triangle is A = (√3 / 4) × s², where s is the length of a side. Substitute s = 8.0 cm (convert to meters: 0.08 m) into the formula to find the area.

Step 3: Rearrange the magnetic flux formula to solve for cos(θ): cos(θ) = Φ / (B × A). Substitute the given values: Φ = 6.0 μWb (convert to webers: 6.0 × 10⁻⁶ Wb), B = 5.0 mT (convert to teslas: 5.0 × 10⁻³ T), and the calculated area A from Step 2.

Step 4: Use the value of cos(θ) obtained in Step 3 to find the angle θ. Take the inverse cosine (arccos) of cos(θ) to determine θ in radians or degrees, depending on the desired unit.

Step 5: Verify the result by checking the units and ensuring the calculated angle is consistent with the physical setup of the problem (e.g., the angle should be between 0° and 90° since the flux is positive).

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Key Concepts

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

Magnetic Flux

Magnetic flux is a measure of the quantity of magnetism, taking into account the strength and the extent of a magnetic field. It is defined as the product of the magnetic field strength (B) and the area (A) through which the field lines pass, adjusted for the angle (θ) between the field and the normal to the surface. The formula is given by Φ = B * A * cos(θ), where Φ is the magnetic flux.

Uniform Magnetic Field

A uniform magnetic field is one in which the magnetic field strength is constant in both magnitude and direction throughout a given region. This means that any magnetic field lines are parallel and evenly spaced, indicating that the force experienced by a charged particle moving through the field will be consistent. In this problem, the uniform magnetic field is specified as 5.0 mT.

Angle Between Magnetic Field and Normal

The angle between the magnetic field and the normal to a surface is crucial for calculating magnetic flux. The normal is an imaginary line perpendicular to the surface, and the angle θ affects how much of the magnetic field passes through the area. In the context of the problem, determining this angle is essential to relate the given magnetic flux to the magnetic field and the area of the triangle.

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Related Practice

Textbook Question

FIGURE EX30.13 shows a 10-cm-diameter loop in three different magnetic fields. The loop's resistance is 0.20 Ω. For each, what are the size and direction of the induced current?

Textbook Question

What is the magnitude of the magnetic flux through the loop shown in FIGURE EX30.5?

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Textbook Question

FIGURE EX30.8 shows a 2.0-cm-diameter solenoid passing through the center of a 6.0-cm-diameter loop. The magnetic field inside the solenoid is 0.20 T. What is the magnitude of the magnetic flux through the loop when it is perpendicular to the solenoid and when it is tilted at a 60° angle?

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Textbook Question

A solenoid is wound as shown in FIGURE EX30.9. Is there an induced current as magnet 2 is moved away from the solenoid? If so, what is the current direction through resistor R?

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Textbook Question

A potential difference of 0.050 V is developed across the 10-cm-long wire of FIGURE EX30.3 as it moves through a magnetic field perpendicular to the figure. What are the strength and direction (in or out) of the magnetic field?

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Textbook Question

INT A 10-cm-long wire is pulled along a U-shaped conducting rail in a perpendicular magnetic field. The total resistance of the wire and rail is 0.20 Ω. Pulling the wire at a steady speed of 4.0 m/s causes 4.0 W of power to be dissipated in the circuit. How big is the pulling force?

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