chapt28 - Chapter 28: Magnetic Fields Problem 1 An electron...

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Chapter 28: Magnetic Fields Problem 1 An electron that has a velocity given inby v = (2.0 × 10 6 i m/s) + (3.0 × 10 6 j m/s) moves through a uniform magnetic field B = 0.030 T i - 0.15 T j . (a) Find the force on the electron due to the magnetic field. (b) Repeat your calculation for a proton having the same velocity. Answer (a) 6.2 × 10 -14 k , N. (b) -6.2 × 10 -14 k , N. Solution (a) The force on a charge q with velocity v in magnetic field B is F = q v × B F = (-1.6 × 10 -19 )*(2.0 × 10 6 i + 3.0 × 10 6 j ) × ( 0.030 i - 0.15 j ) C-m-T/s F = 6.2 × 10 -14 k N. (b) If the charge were a proton instead then q would be +1.6 × 10 -19 and the force would therefore be -6.2 × 10 -14 k N. Problem 5 An electron moves through a uniform magnetic field given by B = B x i + (3.0 B x ) j . At a particular instant, the electron has velocity v = 2.0 m/s i + 4.0 m/s j and the magnetic force acting on it is F = (6.4 × 10 -19 N) k . Find B x . Answer –2.0 T. Solution The force on the electron is F = q v × B F = (-1.6 × 10 -19 C)*( 2.0 m/s i + 4.0 m/s j ) × ( B x i + (3.0 B x ) j ) F = (-3.2 × 10 -19 B z k ) C-m/s Since this force is equal to (6.4 × 10 -19 N) k we obtain -3.2 × 10 -19 B z C-m/s = 6.4 × 10 -19 N. and B z = -2.0 T. Problem 8 An electric field of 1.50 kV/m and a perpendicular magnetic field of 0.400 T act on a moving electron to produce no net force. What is the electron’s speed?
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28-2 Chapter 28: Magnetic Fields Answer 3.75 × 10 3 m/s Solution If the net force is zero and the velocity of the electron is perpendicular to the two fields then the magnitude of the electrostatic force is equal to the magnitude of the magnetic force. qE = qvB v = E/B = (1.50 kV/m) / (0.400 T) = 3.75 × 10 3 m/s. Problem 9 An electron has an initial velocity of (12.0 j + 15.0 k ) km/s and a constant acceleration of (2.00 × 10 -12 m/s 2 ) i in a region in which uniform electric and magnetic fields are present. If B = (400 μ T) i find the electric field E . Answer (-17.4 j + 14.8 k ) V/m Solution Applying the second law to the electron yields q v × B + q E = m a E = m q a v × B E = 9.11 × 10 -31 kg 1.6 × 10 -19 C 2 × 10 12 m/s 2 ( ) j 12 km s j + 15 km s k × 400 μ T ( ) i E = (-17.4 j + 14.8 k ) V/m. Problem 10 A proton travels through uniform magnetic and electric fields. The magnetic field is B = (–2.5 mT) i . At one instant the velocity of the proton is v = (2000 m/s) j m/s. At that instant, what is the magnitude of the net force acting on the proton if the electric field is (a) (4.0 V/m) k ,, (b) -(4.0 V/m) k ,, and (c) (4.0 V/m) i ? Answer (a) (1.44 × 10 -18 N) k ; (b) (6.4 × 10 -19 N) i + (8.0 × 10 -19 N) k . Solution (a) The net force on the proton is F net = q v × B + q E = q( v × B + E ) F net = (1.6 × 10 -19 C)[(2000 m/s) j × (–2.5 mT) i + (4.0 V/m) k ] F net = (1.44 × 10 -18 N) k
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Chapter 28: Magnetic Fields 28-3 (b) The net force on the proton is F net = q( v × B + E ) F net = (1.6 × 10 -19 C)[(2000 m/s) j × (–2.5 mT)
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chapt28 - Chapter 28: Magnetic Fields Problem 1 An electron...

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