(a) Explanation The magnetic field due to the left-hand current to B → L and the due to the right-hand current to B → R . Write the expression for the magnetic field due to a straight wire. B = μ 0 I 2 π r (I) Here, B is the magnetic field, μ 0 is the permeability of free space, I is the current, r is the distance of the point of observation to the straight wire. According to the right hand rule, the magnetic fields are shown in figure 1 . Since the triangle is equilateral, the value of angle θ = 60 ° . The angle ϕ is 90 ° − 60 ° = 30 ° . Due to the symmetry, the vertical components of the magnetic field gets cancel each other and the horizontal components add, so the net magnetic field is directed in the positive x axis. The magnitude of both the magnetic fields are same because same amount of current is passing through each wire. Write the expression for the net magnetic field in the third corner of the triangle. B net = 2 ( μ 0 I 2 π r ) cos ϕ (II) Conclusion: Substitute 4 π × 10 − 7 T ⋅ m / A for μ 0 , 12.0 A for I , 2.50 cm for r , 30 ° for ϕ in equation (II) to find B net (b) To determine The direction of the current flow in the third wire so that the force on it is in the + y direction.

5th Edition

ISBN: 9781260486919

Author: GIAMBATTISTA

Publisher: MCG

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Chapter 19, Problem 111P

(a)

To determine

The magnitude of the magnetic field in the third corner of the triangle.

(b)

To determine

The direction of the current flow in the third wire so that the force on it is in the +y direction.

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Chapter 19, Problem 109P

Chapter 19, Problem 115P

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The magnitude of the magnetic field in the third corner of the triangle.

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The magnitude of the magnetic field in the third corner of the triangle.

The direction of the current flow in the third wire so that the force on it is in the +y direction.

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Chapter 19 Solutions