Two parallel wires, separaed by a distance 20cm, are carring currents (not neces
ID: 1390677 • Letter: T
Question
Two parallel wires, separaed by a distance 20cm, are carring currents (not necessarily the same). One wire carries a current of 15A in the direction shown. Point C is a distance of 10cm to the right of the 15A current and point A is halfway between the two wires.
Additional info :
The magnitude of the B-field at Point C due to the 15A current is 3e-5 T.
The direction of the B field at point C due to the 15 A current is into the page.
If the current on the left is adjusted so that the total magnetic field (due to the fields on both wires) at point C is 6.00e-5 T out of the page , the magnitude of the left hand current is 135.0A.
In which direction would the left hand current have to flow for the total magnetic field at point C to be 6e-5 T out of the page? answer is down
If the currents remain the same as found above, what will be the magnitude of the magneic field for point A ?
answer is 3e-4T
What is the magnitude and direction of the B-field due to the left hand current at the location of the right hand current?
answer is 1.35e-4T(out of page)
Questions;
(vii)What are the magnitude and direction of the force (remember,there is a formula for the force on a current-carrying wire immersed in a magnetic field) on the right hand current due to the legt hand current.
choices: a)4.50e-2N(right) b)2.25e-3N(left) c)4.05e-3N(right) d)6.75e-3N(left)
(viii)If an electron was passing Point moving upward at 1e5 m/s parallel to the wires, what force would that electron experience?(remember you already determined the B-field at point A).
choices:a(7.2e-18N(right) b)1.92e-18N(right) c)9.60e-18N(left) d)4.80e-18N(left)
Explanation / Answer
magnitude and direction of the B-field due to the left hand
current at the location of the right hand current is
BL = 1.35e-4 T (out of page)
the right hand current is in th Bfield of left hand current
the force exerted by B=field of left hand current on the
right hand current F = BL*IR*l
here l = lenght of the second wire is entione;;;;;;''''
F = 2.26e-3 N (left)
viii
when a charges enters a magnetic field with a speed v at an
angle theta with the B-field
F = q*v*B*sintheta
F = 1.6e-19*1e5*3e-4*sin90
F = 4.8e-18 N
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