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kakasveta [241]
2 years ago
5

A current carrying wire of length 50cm is positioned perpendicular to a uniform magnetic field

Physics
1 answer:
RideAnS [48]2 years ago
3 0

A current-carrying wire of length 50 cm is positioned perpendicular to a uniform magnetic field. The magnetic field strength is 0.6 Tesla.

<h3>What is magnetic field?</h3>

The magnetic field is the region of space where an object experiences a magnetic force as it enters the field.

Given is the wire of length 50 cm = 0.5 m and the current is 10.0 A. There is a resultant force of 3.0 N on the wire due to the interaction of the current and field.

The relation between magnetic field strength and current is

F = ILB

Substituting 50 for L, 10 for I and 3 for F, we get the magnetic field strength B.

3 = 10 × 0.5 × B

B= 0.6 Tesla

Therefore, the magnetic field strength is 0.6 Tesla.

Learn more about magnetic field.

#SPJ

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Knowing the speed of the bacteria the uncertainty in its position is
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A plant cell is no longer capable of capturing energy from sunlight and converting it into chemical energy. Which organelle is m
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8 0
1 year ago
Un coche inicia un viaje de 450 km a las ocho de la mañana con una velocidad media de 90 km/h. ¿A qué hora llegará a su destino?
Artyom0805 [142]

Answer:

Llegara a su destino a la 1:00 pm

Explanation:

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5 0
2 years ago
Is the voltage of two identical lamps the same?​
Dahasolnce [82]

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8 0
3 years ago
Read 2 more answers
The notes produced by a tuba range in frequency from approximately 45 Hz to 375 Hz. Find the possible range of wavelengths in ai
taurus [48]

Answer:

The possible range of wavelengths in air produced by the instrument is 7.62 m and 0.914 m respectively.

Explanation:

Given that,

The notes produced by a tuba range in frequency from approximately 45 Hz to 375 Hz.

The speed of sound in air is 343 m/s.

To find,

The wavelength range for the corresponding frequency.

Solution,

The speed of sound is given by the following relation as :

v=f_1\lambda_1

Wavelength for f = 45 Hz is,

\lambda_1=\dfrac{v}{f_1}

\lambda_1=\dfrac{343}{45}=7.62\ m

Wavelength for f = 375 Hz is,

\lambda_2=\dfrac{v}{f_2}

\lambda_2=\dfrac{343}{375}=0.914\ m/s

So, the possible range of wavelengths in air produced by the instrument is 7.62 m and 0.914 m respectively.

6 0
3 years ago
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