Explanation Given data: Diameter of the loop is 10 cm. Magnetic field strength is 0.05 T. Formula used: Write the expression for magnetic flux. Φ m = A B cos θ (1) Here, Φ m is the magnetic flux, A is the area in the magnetic field, B is the magnetic field, and θ is the angle of the flux with respect to the magnetic field. Write the expression for area of the circle. A = π r 2 (2) Here, A is the area of the circle and r is the radius. Explanation: From Figure P25.9, Area of the loop is calculated as follows. Substitute 0.05 m for r in equation (2), A = π × ( 0.05 ) 2 = 7.85 × 10 − 3 m 2 Substitute 0.05 T for B and 7.85 × 10 − 3 m 2 for A in equation (1), Φ m = ( 7.85 × 10 − 3 m 2 ) ( 0.05 T ) cos θ = 3.925 × 10 − 4 × cos θ For 0 ° , Substitute 0 ° for θ in equation (1), Φ m = 3.925 × 10 − 4 × cos 0 ° Wb = 3.925 × 10 − 4 Wb ≅ 3.93 × 10 − 4 Wb For 30 ° , Substitute 0 ° for θ in equation (1), Φ m = 3

College Physics: A Strategic Approach Technology Update, Books a la Carte Plus Mastering Physics with Pearson eText -- Access Card Package (3rd Edition)

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ISBN: 9780134201979

Author: Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field

Publisher: PEARSON

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Chapter 25, Problem 9P

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The magnetic flux through the loop for 0°,30°,60° and 90° .

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Chapter 25, Problem 8P

Chapter 25, Problem 10P

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Chapter 25 Solutions

College Physics: A Strategic Approach Technology Update, Books a la Carte Plus Mastering Physics with Pearson eText -- Access Card Package (3rd Edition)

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The magnetic flux through the loop for 0 ° , 30 ° , 60 ° and 9 0 ° .

Solution Summary: The author explains how the magnetic flux through the loop for 0°,30° and 60° is calculated.

The magnetic flux through the loop for 0°,30°,60° and 90° .

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Chapter 25 Solutions