Lesson_4.1_Homework_and_Solutions

# Lesson_4.1_Homework_and_Solutions - Lesson 4.1 Homework...

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Unformatted text preview: Lesson 4.1 Homework Solutions 1. Find the magnetic force on a proton moving with a velocity of 4.46 Mms-1 i in a magnetic field of 1.75 T k Solution Charge on a proton is q = 1.602 × 10-19 C, Since v and B are at right angles, sin θ = 1 F = qv × B = q v B sin θ = 1.602 × 10-19 × 4.46 × 10 6 ms-1 × 1.75 T = 1.25 × 10-12 N The direction of motion of the proton is the direction of current. Applying Fleming’s left hand rule, we can see that this force is directed in the negative y-direction. F = – 1.25 × 10-12 N j 2. A uniform magnetic field of magnitude 1.48 T is in the positive z-direction. Find the force exerted by this field on a proton if its velocity is (i) v = 2.7 Mms-1 i , (ii) v = 3.7 Mms-1 j , (iii) v = 6.8 Mms-1 k and (iv) v = 4.0 Mms-1 i + 3.0 Mms-1 j Solution (i) F = qv × B = q v B sin θ = 1.602 × 10-19 × 2.7 × 10 6 ms-1 i × 1.48 T k = – 6.4 × 10-13 N j (ii) F = qv × B = q v B sin θ = 1.602 × 10-19 × 3.7 × 10 6 ms-1 j × 1.48 T k = 8.77 × 10-13 N i (iii) Since the direction of motion of the proton is the same as the direction of the filed, there is no force on the proton in this case. (iv) v x produces a force F x and v y produces a force F y given by F x = qv × B = q v B sin θ = 1.602 × 10-19 × 4.0 × 10 6 ms-1 i × 1.48 T k = -9.48 × 10-13 N j F y = qv × B = q v B sin θ = 1.602 × 10-19 × 3.0 × 10 6 ms-1 j × 1.48 T k = 7.11 × 10-13 N i F = 7.11 × 10-13 N i – 9.48 × 10-13 j 3. A straight wire segment 2.0 m long makes an angle of 30 o with a uniform magnetic field of 0.37 T. Find the magnitude of the force on the wire if it carries a current of 2.6 A. Solution I = 2.6 A, B = 0.37 T, θ = 30 o , dℓ = 2.0 m F = Idℓ × B = I dℓ B sin θ = 2.6 A × 2.0 m × 0.37 T × 0.5 = 0.962 N 4. What is the force on an electron with a velocity v = (2 i – 3 j ) Mms-1 in a magnetic field B = (0.8 i + 0.6 j – 0.4 k ) T Solution F = qv × B ( 29...
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Lesson_4.1_Homework_and_Solutions - Lesson 4.1 Homework...

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