A scalloped hammerhead shark swims at a steady speed of 1.0 m/sm/s with its 83...

David Troyer

David Troyer

Answered

2022-01-17

A scalloped hammerhead shark swims at a steady speed of 1.0 m/sm/s with its 83 cm -cm-wide head perpendicular to the earth's 52μT magnetic field.
What is the magnitude of the emf induced between the two sides of the shark's head?
Express your answer using two significant figures.

Answer & Explanation

Matthew Rodriguez

Matthew Rodriguez

Expert

2022-01-18Added 32 answers

Step 1
Given:
The velocity of hammerhead shark (v)=1.0ms
Length of hammerhead (L)=83cm.
the magnetic field of the earth (B)=52μT
the angle between the hammerhead and the earth's magnetic field =90.
Step 2
the magnitude of electromotive force is given by,
Femf=BL×v
Femf=BLvsinθ
Femf=52×106×83×102×1
Femf=4.316×105N
Step 3
Answer:
electromotive force =4.316×105N
Philip Williams

Philip Williams

Expert

2022-01-19Added 39 answers

Given:
- The speed of scalloped hammerhead shark is: va=1.0ms.
- The width of head is: wa=83cm=0.83m.
- The magnetic field is: Ba=52μT=53×106T.
The expression for the change in area through which magnetic field moves is,
Aa=lawa
Here la and wa are change in length and width respectively.
The expression for the speed of scalloped hammerhead shark is,
va=lat
Here is time interval.
The expression for the magnetic flux is,
ϕa=BaAa
The expression for the induced emf by using the Faradays
alenahelenash

alenahelenash

Expert

2022-01-23Added 366 answers

Answer: 7.4×105T. Explanation: Emf induced =BLv Where B is magnetic field , L is length of rod or any other object moving in magnetic field and v is velocity of moving rod. Here B=56×106T L=88×102m v=1.5m/s emf induced =56×106×88×102×1.5 =7.4×105T.

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