The loop in the figure to the right carries a current I in the direction drawn.
ID: 1369339 • Letter: T
Question
The loop in the figure to the right carries a current I in the direction drawn. How is the magnetic field at point A related to the magnetic field at point B? The magnetic field at A is stronger than the magnetic field at B. The magnetic field at A is weaker than the magnetic field at B. The magnetic field at A is the same strength as the magnetic field at B. An electron is accelerated to very near the light (v = 3 times 10^8 m/s) in a particle accelerator It travels down the center of a beam pipe" (as shown to the right) with radius 2.5cm. What is the magnetic field (magnitude and direction, naturally) produced by the electron at the point P at the edge of the beam pipe at the same cross section as the electron? B = 7.7 times 10^-15T into the page B = 7.7 times 10^-15T out of the page B = 6.5 times 10^-9T into the page B = 6.5 times 10^-9T out of page B = 4.9 times 10^-4T into the page. The figure below shows a wire that carries a current I in the direction drawn Four small elements of the current are indicate by tha four rectangles. The distances from the current elements of the picture below are r AP = 2cm, r BP= 1 cm,r CP= 2cm, and r DP = 3cm, and Rank tha strength (magnitude) of the contribution to the magnetic, field by the four segments of the current . BB > BC = BA > BD BB BA = BD BB > BC > BA > BD BC > BA > BB = BDExplanation / Answer
11.1 a)
11.2) a) B = 7.7*10^-7 T into the page
B = (mue/(4*pi)*(q*v/r^2)
= (4*pi*10^-7/(4*pi))*(1.6*10^-19*3*10^8/0.025^2)
= 7.7*10^-15 T
11.3) d) BB > BC > BA > BD
unsing the below equation we can determine above values.
magnetic field due to small segmaent, B = (mue/(4*pi))*(I*dL*sin(theta)/r^2)
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