The magnitude of the magnetic field inside the solenoid is 3.4×10^(-4) T.
To find the answer, we need to know about the magnetic field inside the solenoid.
<h3>What's the expression of magnetic field inside a solenoid?</h3>
- Mathematically, the expression of magnetic field inside the solenoid= μ₀×n×I
- n = no. of turns per unit length and I = current through the solenoid
<h3>What's is the magnetic field inside the solenoid here?</h3>
- Here, n = 290/32cm or 290/0.32 = 906
I= 0.3 A
- So, Magnetic field= 4π×10^(-7)×906×0.3 = 3.4×10^(-4) T.
Thus, we can conclude that the magnitude of the magnetic field inside the solenoid is 3.4×10^(-4) T.
Learn more about the magnetic field inside the solenoid here:
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Answer: a = 1.32 * 10^18m/s² due north
Explanation: The magnitude of the force required to move the electron is given as
F = ma
The force exerted on the charge by the electric field of intensity (E) is given by
F = Eq
Thus
Eq = ma
a = E * q/ m
Where a = acceleration of charge
E = strength of electric field = 7400N/c
q = magnitude of electronic charge = 1.609 * 10^-6c
m = mass of an electronic charge = 9.109 * 10^-31kg
a = 7400 * 1.609 * 10^-16/ 9.109 * 10^-31
a = 11906.6 * 10^-16 / 9.019 * 10^-31
a = 1.19 * 10^-12 / 9.019 * 10^-31
a = 0.132 * 10^19
a = 1.32 * 10^18m/s²
As stated in the question, the direction of the electric field is due north hence, the direction of it force will also be north thus making the electron experience a force due north ( according to Newton second law of motion)
Answer:
A, 0.050 Hz
Explanation:
1) Frequency = speed divided by wavelength
time is 2* 60 = 120 seconds
distance = 6 wave lengths
speed = distance divided by time
speed = 6 wave lengths divided by 120
Hope this helps!
Answer: 2 seconds
Explanation:
Given that,
Time (T) = ?
Charge (Q) = 4 coulombs
current (I) = 2 Amps
Since charge depends on the amount of current flowing through the wire in a given time, hence
Charge = Current x Time
Q = IT
4 coulombs = 2 Amps x Time
Time = 4 coulombs / 2 Amps
Time = 2 seconds
Thus, it takes 2 seconds for the current to flow through the wire
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