course Phy 202 7/28 2:30 pm 030. `Query 30*********************************************
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Given Solution: The force on a current is I * L * B sin(theta) = 150 amps * 160 meters * 5.5 * 10^-5 T * sin(65 deg) = 1.20 amp * m * (N / (amp m) ) = 1.20 Newtons. Note that a Tesla, the unit of magnetic field, has units of Newtons / (amp meter), meaning that a 1 Tesla field acting perpendicular to a 1 amp current in a carrier of length 1 meters produces a force of 1 Newton. The question didn't ask, but be sure you know that the direction of the force is perpendicular to the directions of the current and of the field, as determined by the right-hand rule (fingers in direction of current, hand oriented to 'turn' fingers toward field, thumb in direction of force). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qQuery Principles and General Physics 20.10. Force on electron at 8.75 * 10^5 m/s east in vertical upward magnetic field of .75 T. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: The equation we need to use is F=q*v*B F= (1.6*10^-19 C)*(8.75 *10^5 m/s)*(0.75 T) F= 1.05*10^-13 Cm/s*T F= 1.05*10^-13 N confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 2
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Given Solution: The magnitude of the force on a moving charge, exerted by a magnetic field perpendicular to the direction of motion, is q v B, where q is the charge, v the velocity and B the field. The force in this case is therefore F = q v B = 1.6 * 10^-19 C * 8.75 * 10^5 m/s * .75 T = 1.05 * 10^-13 C m/s * T = 1.05 * 10^-13 N. (units analysis: C m/s * T = C m/s * (N / (amp m) ) = C m/s * (kg m/s^2) / ((C/s) * m), with all units expressed as fundamental units. The C m/s in the numerator 'cancels' with the C m/s in the denominator, leaving kg m/s^2, or Newtons). The direction of the force is determined by the right-hand rule (q v X B) with the fingers in the direction of the vector q v, with the hand oriented to turn the fingers toward the direction of B. The charge q of the electron is negative, so q v will be in the direction opposite v, to the west. In order for the fingers to 'turn' qv toward B, the palm will therefore be facing upward, the fingers toward the west, so that the thumb will be pointing to the north. The force is therefore directed to the north. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 2 ********************************************* Question: `qWhat is the radius of orbit for a proton with kinetic energy 5.4 MeV? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: If we double the kinetic energy, we will increase the velocity of that proton by sqrt(2). So the radius of the orbit is also increased by sqrt(2) R= .096 m confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 2
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Given Solution: ** Doubling the KE of the proton increases velocity by factor sqrt(2) and therefore increases the radius of the orbit by the same factor. We end up with a radius of about .096 m. ** STUDENT QUESTION what numbers were used to find this? INSTRUCTOR RESPONSE In Problem 27.60, above, we found the radius of orbit for a proton with kinetic energy 2.7 MeV. Here we are finding the radius for a proton with twice the KE. We could do this in the same manner as before, and we would get the same result. However thinking in terms of the proportionality, as is done here, is both more efficient and more instructive. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 "