Calculate the work done in lifting a 500-N
1 answer:
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Correct question:
A solenoid of length 0.35 m and diameter 0.040 m carries a current of 5.0 A through its windings. If the magnetic field in the center of the solenoid is 2.8 x 10⁻² T, what is the number of turns per meter for this solenoid?
Answer:
the number of turns per meter for the solenoid is 4.5 x 10³ turns/m.
Explanation:
Given;
length of solenoid, L= 0.35 m
diameter of the solenoid, d = 0.04 m
current through the solenoid, I = 5.0 A
magnetic field in the center of the solenoid, 2.8 x 10⁻² T
The number of turns per meter for the solenoid is calculated as follows;
Therefore, the number of turns per meter for the solenoid is 4.5 x 10³ turns/m.
Answer:
The change on the second particle is .
Explanation:
The period of revolution of the particle in the magnetic field is given by the formula as follows :
It is given that the magnetic field is uniform. The mass of the second particle is the same as that of a proton but thecharge of this particle is different from that of a proton.
If both particles take the same amount of time to go once around their respective circles. So,
So, the change on the second particle is .