Answer:
The number of turns in the solenoid is 230.
Explanation:
Given that,
Rate of change of current, 
Induced emf, 
Current, I = 1.5 A
Magnetic flux, 
The induced emf through the solenoid is given by :
or
........(1)
The self inductance of the solenoid is given by :
.........(2)
From equation (1) and (2) we get :
N is the number of turns in the solenoid
N = 229.28 turns
or
N = 230 turns
So, the number of turns in the solenoid is 230. Hence, this is the required solution.
Answer:
2.6h
Explanation:
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Answer:
the magnitude of a uniform electric field that will stop these protons in a distance of 2 m is 10143.57 V/m or 1.01 × 10⁴ V/m
Explanation:
Given the data in the question;
Kinetic energy of each proton that makes up the beam = 3.25 × 10⁻¹⁵ J
Mass of proton = 1.673 × 10⁻²⁷ kg
Charge of proton = 1.602 × 10⁻¹⁹ C
distance d = 2 m
we know that
Kinetic Energy = Charge of proton × Potential difference ΔV
so
Potential difference ΔV = Kinetic Energy / Charge of proton
we substitute
Potential difference ΔV = ( 3.25 × 10⁻¹⁵ ) / ( 1.602 × 10⁻¹⁹ )
Potential difference ΔV = 20287.14 V
Now, the magnitude of a uniform electric field that will stop these protons in a distance of 2 m will be;
E = Potential difference ΔV / distance d
we substitute
E = 20287.14 V / 2 m
E = 10143.57 V/m or 1.01 × 10⁴ V/m
Therefore, the magnitude of a uniform electric field that will stop these protons in a distance of 2 m is 10143.57 V/m or 1.01 × 10⁴ V/m
