A long, thin solenoid has 450 turns per meter and a radius of 1.17 cm. The current in the solenoid is increasing at a uniform rate did. The magnitude of the induced electric field at a point which is near the center of the solenoid and a distance of 3.45 cm from its axis is 8.20×10−6 V/m.Calculate di/dt
di/dt = _________.
Answers
The value of di/dt from the given values of the solenoid electric field is;
di/dt = 7.415 A/s
We are given;
Number of turns; N = 450 per m
Radius; r = 1.17 cm = 0.0117 m
Electric Field; E = 8.2 × 10⁻⁶ V/m
Position of electric field; r' = 3.45 cm = 0.0345 m
According to Gauss's law of electric field;
∫| E*dl | = |-d∅/dt |
Now, ∅ = BA = μ₀niA
where;
n is number of turns
i is current
A is Area
μ₀ = 4π × 10⁻⁷ H/m
Thus;
E(2πr') = (d/dt)(μ₀niA) (negative sign is gone from the right hand side because we are dealing with magnitude)
Since we are looking for di/dt, then we have;
E(2πr') = (di/dt)(μ₀nA)
Making di/dt the subject of the formula gives;
di/dt = E(2πr')/(μ₀nA)
Plugging in the relevant values gives us;
di/dt = (8.2 × 10⁻⁶ × 2 × π × 0.0345)/(4π × 10⁻⁷ × 450 × π × 0.0117²)
di/dt = 7.415 A/s
Read more at; brainly.com/question/14003638
Answer:
Explanation:
From the question we are told that
The number of turns is
The radius is
The position from the center consider is x = 3.45 cm = 0.0345 m
The induced emf is
Generally according to Gauss law
=>
Where A is the cross-sectional area of the solenoid which is mathematically represented as
=>
=> ggl;
Here is the permeability of free space with value
=>
=>
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Unknown
Force applied = ?
Solution:
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