Explanation Given data: B L = 2 T . B R = 1 T . Length of magnetic field on left is 20 cm. Length of magnetic field on right is 20 cm Breadth of magnetic field is 20 cm. Formula used: Consider the expression of magnetic flux. Φ = B A cos θ (1) Here, B is the magnetic field strength, A is the area, and θ is the angle between magnetic field and area. Explanation: Refer to Figure P25.10 in the textbook. Since the loop is in the rectangular shape, the area of the magnetic field present on left side is, A = Length ×Breadth (2) Substitute 20 cm for Length and 20 cm for Breadth in equation (2), A L = 20 cm ×20 cm = 0 .2 m × 0 .2 m = 0 .04 m 2 Consider the area to be into the page. So, the angle made by B L with area is 0 ° because B L is also into the page. Magnetic flux on left side is calculated as follows. Substitute 2 T for B , 0 ° for θ , and 0 .04 m 2 for A in equation (1), Φ L = 2 T × 0 .04 m 2 × cos 0 ° = 0.08 Wb Area of the magnetic field present on right side is calculated as follows. Substitute 20 cm for Length and 20 cm for Breadth in equation (2), A R = 20 cm ×20 cm = 0

College Physics: A Strategic Approach (3rd Edition)

3rd Edition

ISBN: 9780321879721

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

Publisher: PEARSON

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Chapter 25, Problem 10P

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The magnetic flux through the loop.

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Chapter 25, Problem 9P

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Chapter 25 Solutions

College Physics: A Strategic Approach (3rd Edition)

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The magnetic flux through the loop.

Solution Summary: The author explains that the magnetic flux through the loop is 0.040 Wb.

The magnetic flux through the loop.

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Chapter 25 Solutions