Explanation Given: The given field strength H of a magnet of length 2 L on a particle r units from the centre of the magnet is: H = 2 m L ( r 2 + L 2 ) 3 2 Where ± m are the poles of the magnet. Formula used: ∫ sin x d x = − cos x + c ∫ cos x d x = sin x + c Calculation: Consider the following field strength function H = 2 m L ( r 2 + L 2 ) 3 2 ; As it moves from 0 to R units from the centre, so, the average field strength of magnet is: H arg = 1 R ∫ 0 R 2 m L ( r 2 + L 2 ) 3 2 Now rewrite the integral as; H arg = 1 R ∫ 0 R 2 m L ( r 2 + L 2 ) 3 d r ……..(1) Consider the following triangle, We get, tan θ = r L r = L tan θ And, r 2 + L 2 = L sec θ Now differentiate with respect to θ ; d r = L sec 2 θ d θ Substitute these values in equation (1); H arg = 1 R ∫ 0 R 2 m L ( r 2 + L 2 ) 3 d r H arg = 1 R ∫ 0 R 2 m L ( L sec θ ) 3 ( L sec 2 θ ) d θ H arg

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Chapter 8.4, Problem 68E

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To calculate: The average field strength H of the magnet as it moves from0 to R units from the centre.

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Chapter 8.4, Problem 58E

Chapter 8.4, Problem 69E

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The average field strength H of the magnet as it moves from0 to R units from the centre.

Solution Summary: The author calculates the average field strength of the magnet as it moves from 0 to R units from the centre.

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