Technician A says a short to voltage may or may not blow a fuse and may or may not affect more than one circuit. Technician B sa
1 answer:
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Answer:
The magnitude of the magnetic field vector is 1.91T and is directed towards the east.
The steps to the solution can be found in the attachment below.
Explanation:
For the charge to remain in the the earth' gravitational field the magnetic force on the charge must be equal to the earth's gravitational force on the charge and must act opposite the direction of the earth's gravitational force.
Fm = Fg
qvBSin(theta) = mg
Where q = magnitude of charge
v = magnitude of the velocity vector = 4 x10^4 m/s
B = magnitude of the magnetic field vector
theta = the angle between the magnetic field and velocity vectors = 90°
m = mass of the charge = 0.195g
g = acceleration due to gravity =9.8m/s²
On substituting the respective values of all variables in the equation (1) above
B = 1.91T
The direction of the magnetic field vector was found by the application of the right hand rule: if the thumb is pointed in the direction of the magnetic force and the index finger is pointed in the direction of the velocity vector, the middle finger points in the direction of the magnetic field.
Below is the step by step procedure to the solution.
The self-inductance of a solenoid is given by:
where
is the vacuum permeability
N is the number of turns
A is the cross-sectional area of the solenoid
l is the length of the solenoid
For the solenoid in our problem, N=3000, l=70.0 cm=0.70 m and the self-inductance is L=25.0 mH=0.025 H, therefore the cross-sectional area is
And since the area is related to the radius by
The radius of the solenoid is
where
N is the number of turns
A is the cross-sectional area of the solenoid
l is the length of the solenoid
For the solenoid in our problem, N=3000, l=70.0 cm=0.70 m and the self-inductance is L=25.0 mH=0.025 H, therefore the cross-sectional area is
And since the area is related to the radius by
The radius of the solenoid is