What is the direction of the magnetic field hatwide b

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What is the direction of the magnetic field hatwide B P at point P due to the upward current in the left-hand side of the square wire? 1. hatwide B is up the page. 2. hatwide B is out of the page. 3. hatwide B is zero. 4. hatwide B is to the left. 5. hatwide B is to the right. See Part 2. 011 10.0points An electron is in a uniform magnetic field B that is directed out of the plane of the page, as shown. v e B B B B 6. hatwide B is down the page. 7. hatwide B is into the page. correct Explanation: The direction of the magnetic field due to a current element is determined by the cross product in the definition of the magnetic field vector B = μ 0 I 4 π integraldisplay dvectors × ˆ r r 2 . For the present case the right hand rule gives the direction of the magnetic field as into the page or in for short. 010(part3of3)10.0points What is the direction of the magnetic field hatwide B P at point P due the left-to-right current in the top side of the square wire? 1. hatwide B is zero. 2. hatwide B is into the page. correct 3. hatwide B is to the right. 4. hatwide B is to the left. 5. hatwide B is down the page. 6. hatwide B is up the page. 7. hatwide B is out of the page. Explanation:
dong (zd742) – HW23- Magnetism/EM – mcmeans – (00004) 4 When the electron is moving in the plane of the page in the direction indicated by the arrow, the force on the electron is directed 1. into the page. 2. toward the top of the page. 3. toward the right 4. toward the bottom of the page. correct 5. out of the page. 6. toward the left q v B sin θ = (2 . 8 × 10 8 C) (69 m / s) (4 . 6 × 10 5 T) × sin (33 ) = 4 . 84032 × 10 11 N . (The Lorentz force is the name given to the force acting on a charged particle moving in a magnetic field.) 013(part1of2)10.0points A proton moves perpendicularly to a uni- form magnetic field B with a speed of 6 × 10 7 m / s and experiences an acceleration of 1 × 10 13 m / s 2 in the positive x direction when its velocity is in the positive z direction. The mass of a proton is 1 . 673 × 10 27 kg. Find the magnitude of the field. 012 10.0points A duck flying due north at 69 m / s passes over = q v B sin θ = (2 . 8 × 10 8 C) (69 m / s) (4 . 6 × 10 5 T) × sin (33 ) = 4 . 84032 × 10 11 N . (The Lorentz force is the name given to the force acting on a charged particle moving in a magnetic field.) 013(part1of2)10.0points A proton moves perpendicularly to a uni- form magnetic field B with a speed of 6 × 10 7 m / s and experiences an acceleration of 1 × 10 13 m / s 2 in the positive x direction when its velocity is in the positive z direction. The mass of a proton is 1 . 673 × 10 27 kg. Find the magnitude of the field.
Atlanta, where the Earth’s magnetic field is 4 . 6 × 10 5 T in a direction 33 below the horizontal line running north and south. If the duck has a net positive charge of 2 . 8 × 10 8 C, what is the magnitude of the magnetic force acting on it? F = F magnetic m a = q v B . Thus B = m a q v ( 1 . 673 × 10 27 kg ) ( 1 × 10 13 m / s 2 )
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