A thin ring of radius 73 cm carries a positive charge of 610 nC uniformly distributed over it. A point charge q is placed at the
1 answer:
Answer:
q = - 93.334 nC
Explanation:
GIVEN DATA:
Radius of ring 73 cm
charge on ring 610 nC
ELECTRIC FIELD p FROM CENTRE IS AT 70 CM
E = 2000 N/C
Electric field due tor ring is guiven as
E1 = 3714.672 N/C
electric field due to point charge q
now the eelctric charge at point P is
E = E1 + E2
solving for q
q = - 93.334 nC
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(a) 34 V
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(1)
where
is the variation of magnetic flux through the coil
is the time interval
We need to find the magnetic flux before and after. The magnetic flux is given by:
where
B is the magnetic field intensity
A is the area of the coil
The radius of the coil is r = 12.0 cm = 0.12 m, so its area is
At the beginning, the magnetic field is
so the flux is
while after the removal of the coil, the magnetic field is zero, so the flux is also zero:
so the variation of magnetic flux is
And substituting into (1) we find the average emf in the coil
(b) Counterclockwise
In order to understand the direction of the induced current, we have to keep in mind the negative sign in Lenz's law (1), which tells that the direction of the induced current must be such that the magnetic field produced by this current opposes the variation of magnetic flux in the coil.
In this situation, the magnetic flux through the coil is decreasing, since the coil is removed from the field. So, the induced current must be such that it produces a magnetic field whose direction is the same as the direction of the external magnetic field, which is upward along the positive z-direction.
Looking down from above and using the right-hand rule on the loop (thumb: direction of the current, other fingers wrapped: direction of magnetic field), we see that in order to produce at the center of the coil a magnetic field which is along positive z-direction, the induced current must be counterclockwise.