(Be clear and concise. Typing work is strongly suggested. Will thumb up based on
ID: 1421083 • Letter: #
Question
(Be clear and concise. Typing work is strongly suggested. Will thumb up based on the quality of the answer given.) (Please answer question 1Aand 1B) A charged particle (mass = 4.0 mu g, charge = 5.0 mu C) moves in a region where the only force on it is magnetic. What is the magnitude of the acceleration of the particle at a point where the speed of the particle is 5.0 km/s, the magnitude of the magnetic field is 8.0 mT, and the angle between the direction of the magnetic field and the velocity of the particle is 60 degree? (Give your answer in km/s^2) A straight wire of length 70 cm carries a current of 50 A and makes an angle of 60 degree with a uniform magnetic field. If the force on the wire is 1.0 N what is the magnitude of B? (Give your answer in mT. 1mT=10^-3T)Explanation / Answer
(a),F = q(v×B) where × is the cross product
F = qvBsin() where is angle between velocity and magnetic field vector.
F = 5*10-6 * 5000*8*10-3sin(60)
F = ma =173.2 N
a = 173.2/4*10-3 = 43300 m/s2
(b) F = I(L×B)
1 = 50*0.7*B*sin(60)
B = 32.99 mT
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