An electron is accelerated through 2400 V from rest and then enters a region whe

Question

An electron is accelerated through 2400 V from rest and then enters a region where there is a uniform 1.40 T magnetic field. What are the maximum and minimum magnitudes of the magnetic force acting on this electron l Verizon LTE 5:14 PM Personal Hotspot:1 Connecti An electron is accelerated through 2400 V from rest and then enters a region where there is a uniform 1.40 T magnetic field. What are the maximum and minimum magnitudes of the magnetic force acting on this electron? (a) maximum 271744x Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. N (b) minimum Sodium ions (Nat) move at 0.849 m/s through a bloodstream in the arm of a person standing near a large magnet. The magnetic field has a strength of 0.253 T and makes an angle of 50.0° with the motion of the sodium ions. The arm contains 115 cm3 of blood with 4.00 x 1020 Nat ions per cubic centimeter. If no other ions were present in the arm, what would be the magnetic force on the arm? A proton moves perpendicularly to a uniform magnetic field s at 2.0 x 107 m/s and exhibits an acceleration of 5.0 x 1013 m/s2 in the +x-direction when its velocity is in the +z-direction. Determine the magnitude and direction of the field. A wire carries a current of 14.0 A in a direction that makes an angle of 29.0° with the direction of a magnetic field of strength 0.500 T. Find the magnetic force on a 4.00 m length of the wire. A wire 2.80 m in length carries a current of 5.60 A in a region where a uniform magnetic field has a magnitude of 0.330 T. Calculate the magnitude of the magnetic force on the wire for the following angles between the magnetic field and the current. (a) 60.0

Explanation / Answer

3. KE = q delta(V)

(9.109 x 10^-31) v^2 /2 = (1.6 x 10^-19) (2400)

v = 2.90 x 10^7 m/s

F = q (v X B ) = q v B sin(theta)

(A)maximum when theta = 90 deg  

F = q v B =6.50 x 10^-12 N

(B) minimum when theta = 0

F = 0
--------------------------

4. v = 0.849 m/s

B = 0.253 T

theta = 50 deg

number of ions = (4 x 10^20) (115) = 4.6 x 10^22 ions

q = (1.6 x 10^-19)(4.6 x 10^22)

q = 7360 C

F = q v B sin(theta)

= 7360 x 0.849 x 0.253 x sin50

= 1211 N

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