6.67 points Save Answer Question 4 An electron moves in a region where the magne
ID: 1792578 • Letter: 6
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6.67 points Save Answer Question 4 An electron moves in a region where the magnetic field is uniform and has a magnitude of S0 pT. The electron follows a helical path whieh has a piteh of 9.0 mm and a radius of 2.0 mm. What is the speed of this electron as it moves in this region? 0.35 km/s b. 8.0 km/s C. 20 kms d.48 km/s e, 28 kms 6.67 points Save Answer Question5 Two long straight parallel wires separated by a distance of 20 cm carry currents of 30 A and 40 A in opposite directions. What is the magnitude of the resulting magnetic field at a point that is 15 cm from the wire carrying the 30-A current and 25 cm from the other wire? d.72T 6.67 points Save Answer Question 6 The figure shows a cross section of three parallel wires each carrying a curent of 5.0 A out of the paper. If the distance R-6.0 mm, what is the magnitude of the magnetic force on a 2.0-m length of any one of the wires? a. 2.9 mN b.2.2 mN c, 1.7 mN d.2.5 mN e. 3.3mNExplanation / Answer
4)
period of orbit is
T = 2pim/qB
= (2*3.14*9.1*10^-31)/(1.6*10^-19*80*10^-6)
= 4.46*10^-7 s
pitch = v T
v= pitch/T = 9.0 * 10^-3 / 4.46*10^-7 s = 20 km/s
5)
the net magnetic field
B = sqrt B1^2 + B2^2 + 2 B1 B2 cos theta
since theta = 90 + tehta
cos ( 90 + theta) = sin thea
= sqrt ( uo i1 2pi r1 )^2 + ( uo I2/ 2pi r2)^2 - 2( uo i1 2pi r1) ( uo I2/ 2pi r2)sin theta
= sqrt ( 2 * 10^-7 ( 30)/ 0.15)^2 + ( 2 * 10^-7 ( 40)/0.25)^2 - 2( 2 * 10^-7 ( 30)/ 0.15) (2 * 10^-7 ( 40)/ 0.25 * (0.15/0.25)
= 33 uT
The magentic force acting on a wire is
F /L = uo I^2/ 2 pi R
Fx = uo I^2 L/ 2pi R * ( 1 + cos 60) ( -i)
= 4 pi * 10^-7 5^2 2 / 2pi ( 6.0 * 10^-3) * ( 1+ 0.5) ( -i)
=25 * 10^-4 N ( -i)
Fy = uo I^2 L/ 2pi R * sin 60 ( j)
=14.43 * 10^-4 N j
F = sqrt Fx^2 + Fy ^2
= sqrt ( 25 * 10^-4)^2 + ( 5.77 * 10^-4)^2
=2.9 mN
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