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shepuryov [24]
3 years ago
7

As a diligent physics student, you carry out physics experiments at every opportunity. At this opportunity, you carry a 1.01 m l

ong rod as you jog at 3.47 m/s , holding the rod perpendicular to your direction of motion.
(a) What is the strength of the magnetic field that is perpendicular to both the rod and your direction of motion and that induces an EMF of 0.265 mV across the rod? Express the answer in milliteslas.
Physics
1 answer:
Valentin [98]3 years ago
6 0

Answer:

The strength of the magnetic field is 75.6\ \mu T.

Explanation:

Given that,

Length of the rod, L = 1.01 m

Speed with which the rod is moving, v = 3.47 m/s

We need to find the strength of the magnetic field that is perpendicular to both the rod and your direction of motion and that induces an EMF of 0.265 mV across the rod. When the rod is moving with some speed, an emf gets induced and it is given by :

\epsilon=Blv

B is magnetic field

B=\dfrac{\epsilon}{lv}\\\\B=\dfrac{0.265\times 10^{-3}}{1.01\times 3.47}\\\\B=75.6\ \mu T

So, the strength of the magnetic field is 75.6\ \mu T.

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Answer:

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            DE = 12.094 eV

let's reduce the energy to the SI system

            DE = 12.094 eV (1.6 10⁻¹⁹ J / 1 ev) = 10.35 10⁻¹⁹ J

let's find the wavelength is this energy, let's use Planck's equation to find the frequency

            E = h f

             f = E / h

            f = 19.35 10⁻¹⁹ / 6.63 10⁻³⁴

            f = 2.9186 10¹⁵ Hz

now we can look up the wavelength

           c = λ f

           λ = c / f

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