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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 = 1 F = qv B = q v B sin = 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 Flemings 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 = 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 = 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 = 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 = 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, = 30 o , d = 2.0 m F = Id B = I d B sin = 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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