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3 years ago

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As a diligent physics student, you carry out physics experiments at every opportunity. At this opportunity, you carry a 1.33 m l

ong rod as you jog at 3.19 m/s , holding the rod perpendicular to your direction of motion. 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 0.263 mV across the rod

2 answers:

deff fn [24]3 years ago

8 0

Answer: 62 μT

Explanation:

Given

Length of rod, l = 1.33 m

Velocity of rod, v = 3.19 m/s

Induced emf, e = 0.263*10^-3 V

Using Faraday's law, the induced emf of a rod can be gotten by the formula

e = blv where,

e = induced emf of the rod

b = magnetic field of the rod

l = length of the rod

v = velocity of the rod. On substituting, we have

0.263*10^-3 = b * 1.33 * 3.19

0.263*10^-3 = b * 4.2427

b = 0.263*10^-3 / 4.2427

b = 0.0000620 T

b = 62 μT

Thus, the strength of the magnetic field is 62 μT

tamaranim1 [39]3 years ago

4 0

Answer:

The strength of the magnetic field is 0.062 mT

Explanation:

Given:

ε = 0.263 mV

L = 1.33 m

v = 3.19

The induced emf in the rod is equal to:

$\epsilon =BLvsin90\\B=\frac{\epsilon }{Lv} =\frac{0.263}{1.33*3.19} =0.062mT$

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