A long solenoid with 8.22 turns/cm and a radius of 7.00 cm carries a current of 19.4 mA. A current of 3.59 A exists in a straight conductor located along the central axis of the solenoid. (a) At what radial distance from the axis in centimeters will the direction of the resulting magnetic field be at 49.7° to the axial direction? (b) What is the magnitude of the magnetic field there?
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Ответ:
a. 3.039cm
b.magnetic field is![B=2.958\times10^{-5}T](/tpl/images/0548/5101/813fe.png)
Explanation:
Direction of the solenoid magnetic field is along the axis of the solenoid. and magnetic field due to the wire perpendicular to that due to the solenoid.. Magnetic field at r is given by:
Angle of net magnetic field from axial direction is given by:
Field due to solenoid:
Field due to wire:
Therefore, r:
Hence, the radial distance is 3.039cm
b.The magnetic field strength is given by:
Hence, the magnetic field is![B=2.958\times10^{-5}T](/tpl/images/0548/5101/813fe.png)
Ответ:
I got you.. i'm in middle school and had that same question.
Explanation:
Refer to the diagram shown below.
The vertical distance traveled is
s = 25 m
The initial vertical launch velocity is zero.
Therefore
s = (1/2)*g*t²
where g = 9.8 m/s²
t = the time of flight, s
That is,
0.5*9.8*t² = 25
t² = 25/4.9 = 5.102
t = 2.26 s
2.26 s