Physics 205B Quiz 6, spring semester 2014
Cuesta College, San Luis Obispo, CA
A q = +4.0 µC charge moves at a speed of 4.0×104 m/s through a uniform magnetic field with magnitude 4.50 T. The magnitude of the magnetic force on the charge is:
(A) 0 N.
(B) 0.51 N.
(C) 0.72 N.
(D) 1.0 N.
Correct answer (highlight to unhide): (B)
The magnitude of the force of this magnetic field on this moving charge is given by:
FB = |q·v·B·sinθ|,
where the angle θ between the charge's velocity, and the magnetic field is 45°, such that:
FB = |(+4.0×10–6 C)·(4.0×104 m/s)·(4.50 T)·sin(45°)|,
FB = 0.5091168825... N,
or to two significant figures, the magnitude of the magnetic force on the charge is 0.51 N.
(Response (C) is just |q·v·B|; response (D) is |q·v·B/sin(45°)|.)
Sections 30882, 30883
Exam code: quiz06Lp4s
(A) : 0 students
(B) : 24 students
(C) : 13 students
(D) : 0 students
Success level: 65%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.55
20180408
Physics quiz question: magnitude of magnetic force on moving charge
Labels:
charge,
current,
force,
magnet,
magnetic field,
physics multiple-choice question,
RHR,
vectors,
velocity
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