A 15.Zero-cm diameter coil of twine is inner a 0.521-T magnetic field. What Emf is precipitated
when it remains desk bound for 15.0 s? - ANS-No Emf is triggered on this twine on account that
there is no alternate to create the Emf since it remains desk bound.
A charged item actions at 475 m/s to the South thru a magnetic subject of one.42 T, that's
directed to the East. If the price of the object is +12.5 nC, what is the magnitude and direction of
the magnetic force appearing at the item? - ANS-Given: v, B, q wherein magnetic subject
directed east and object transferringVBsin(90)
3 sig figs answer in Mew Newtons
RHR states pressure is Up (?)
A long instantly wire includes a current of 760 mA directly down. The area this is created with
the aid of this present day is being tested at factor A, which is to the North of the twine. The
energy of the magnetic area is zero.357 µT at point A. How a long way away from the wire is
point A, and in which direction is the magnetic subject? - ANS-r= (Mew)oI/2piB wherein (Mew)o
= 4pix10^-7N/A^2 and B = 3.57x10^-7 T
Answer in 3 sig figs (zero.43) solution in m
A square loop of twine this is 5.Zero cm throughout actions right into a place of area in which
the magnetic subject has a value of zero.20 T, as shown in Figure nine.T.1. The loop is pulled by
using some unseen pressure at a consistent pace of 80.0 m/s, as proven within the diagram.
Answer the following questions on this scenario.
A. Calculate the Emf brought on within the loop because it actions from a location where there's
no magnetic field into the magnetic discipline proven.
B. In which course is the Emf prompted inside the loop as it enters the magnetic subject?
(clockwise or counter-clockwise)
c. In which direction does conventional modern flow through the top horizontal section of the
loop? (proper to left or left to proper)
d. If the overall resistance of the loop of cord is 1.50 , what is the amount of present day flowing
via the loop as it enters the sector?
E. Calculate the importance of the force exerted at the top horizontal section of cord while it
actions in the - ANS-a) Use EMF =Blv Answer in volts
b) RHR clockwise
c) Left to Right
d) I=V/R Answer in Amps
e) F =BIL Answer in Newtons
f) RHR Down the page
g) region = side^2 solution in m^2
magnetic flux = ABcos(zero)
solution in wb