A straight 1.0 m long wire is carrying a current. the wire is placed perpendicular to a magnetic field of strength 0.20 t. if th
1 answer:
A current carrying wire placed perpendicular to a magnetic field experiences a force equal to
where
I is the current in the wire
L is the wire length
B is the intensity of the magnetic field
In our problem, the length of the wire is L=1.0 m, the magnetic field strength is B=0.20 T and the force exerted on the wire is F=0.60 N. If we re-arrange the equation and we plug these numbers into it, we find the current in the wire:
where
I is the current in the wire
L is the wire length
B is the intensity of the magnetic field
In our problem, the length of the wire is L=1.0 m, the magnetic field strength is B=0.20 T and the force exerted on the wire is F=0.60 N. If we re-arrange the equation and we plug these numbers into it, we find the current in the wire:
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Answer:
(a) if she increases the tension in the string is increased by 15%, the fundamental frequency will be increased to 740.6 Hz
(b) If she decrease the length of the the string by one-third the fundamental frequency will be increased to 840 Hz
Explanation:
(a) The fundamental, f₁, frequency is given as follows;
Where;
T = The tension in the string
μ = The linear density of the string
L = The length of the string
f₁ = The fundamental frequency = 560 Hz
If the tension in the string is increased by 15%, we will have;
Therefore, if the tension in the string is increased by 15%, the fundamental frequency will be increased by a fraction of 0.3225 or 32.25% to 740.6 Hz
(b) When the string length is decreased by one-third, we have;
The new length of the string, = 2/3·L
The value of the fundamental frequency will then be given as follows;
When the string length is reduced by one-third, the fundamental frequency increases to one-half or 50% to 840 Hz.