i 65) A long straight wire on the s-axis carries a current of s 0 A in the +e di
ID: 1631489 • Letter: I
Question
i 65) A long straight wire on the s-axis carries a current of s 0 A in the +e direction (out of the . A circular loop of radius 10 cm lies in the xy-plane and carries a 3.0-A current, as shown Pint P, at the center of the loop, is 25 em from the s-axis. An electron is avea lhe electron? (e-1.60 × 10-19.0- 10-7 T-mA) the figure. Pi direction. What is the y component of the projected from P with a velocity of 3.0 106m/s in the x magnetic force on Ehe electron? (e-1 60-10-19 AN 4x what h the ycomponent ofte, 3.0 A 8.0A 25 cm·10cm : A)-9.0 × 10-18 N B) 9.0 x 10-18 N C)-4.5 x 10-18 N D) 4.5 × 10-18 N E) zero Answer: A Var: I 66) An ideal solenoid 20 cm long is wound with 5000 turns of very thin wire. What strength magnetic field is produced at the center of the solenoid when a current of 10 A fnlows through in (0-4 × 10-7 T . mA) A) 0.0063 T B) 0.20 T C) 3.2 T D) 4.8 T E) 0.31 T Answer: E Var: 1 67) An ideal solenoid having a coil density of 5000 turns per mfeter is 10 cm long and carries a current of 4.0 A. What is the strength of the magnetic field at its center? A) 3.1 mT B) 6.2 mT C) 13 mT D) 25 mT Answer: D Var: I 47 Copyright © 2017 Pearson Education, Inc.Explanation / Answer
To answer this question you use the following equation for the Lorentz Force:
F = q(E + v × B)
Where q is the charge of the particle in question, E is the electric field (this is actually a vector quantity), v is the velocity of the particle (also a vector) and B is the magnetic field (also a vector). × Indicates that we are taking the cross product. Since we have no electric field our equation for the Lorentz Force simplifies to:
F = qv × B
We have a magnetic field created by the current in the wires thus we cannot simplify the equation any further, however this is not entirely true, we can get rid of the cross product as the magnetic field from the long wire along the z-axis is parallel to the velocity and thus gives zero force. The magnetic field in the center of the loop is perpendicular to the velocity and thus contributes with a force equal to qvB where we have eliminated the cross product replacing it by the right hand rule (this is only possible when the vectors are perpendicular at 90 degrees to each other).
In general the magnetic field in the center of a current carrying loop is given by:
B = µI / 2r
Where µ is the permeability of free space (just a constant), I is the current and r is the radius of the loop.
B is strictly speaking a vector quantity, but I have left out all the nastiest such as cross products etc. Just use the right hand rule to find its direction. The magnetic field will thus be perpendicular to the velocity of the particle (electron). Before calculating there is one more piece of information which we really need and that is the direction of the current within the loop. As you have not given the direction of this current I can only give you the correct magnitude of the y component of the force. I am not able to tell you whether it is negative or positive:
The Lorentz Force becomes:
F = qvB
= [(-1.6 x 10^-19)(3 x 10^6)(4pi x 10^-7)(3)]/(2 x 0.1) = -9.0 x 10^-18 N (ans) ->option A
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