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3 years ago

10

In the rainy season, the Amazon flows fast and runs deep. In one location, the river is 22 m deep and moves at a speed of 4.4 m/

s toward the east. The earth's 50 μT magnetic field is parallel to the ground and directed northward. If the bottom of the river is at 0V, what is the potential (magnitude and sign) at the surface?

2 answers:

andreev551 [17]3 years ago

8 0

Given Information:

depth = L = 22 m

Speed = v = 4.4 m/s

Magnetic field = B = 50 μT

Required Information:

Potential difference = E = ?

Answer:

Potential difference = 4.84 mV

Explanation:

The potential difference can be found using

E = BvL

Where v is the speed of flowing river, B is the magnetic field and L is the depth of river .

E = 50x10⁻⁶*4.4*22

E = 0.00484 V

or

E = 4.84 mV

The direction of current can found using right hand rule, thumb upward indicating force, forefinger indicating magnetic field and curled fingers indicating current.

Since the potential at bottom of the river is 0V then the potential at top of the surface would be +4.84 mV.

Anna35 [415]3 years ago

5 0

Answer:

V = 4.84×10-³V

Explanation:

V = E×d

E = Bv

Given

B = 50μT = 50×10-⁶T = magnetic field

v = 4.4m/s = speed

E = Bv = 50×10-⁶×4.4 = Electric field strength

E = 2.2×10-⁴ V/m

d = 22m = distance or separation

V = Ed =2.2×10-⁴ × 22

V = 4.84×10-³V = potential

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