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.

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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&#039;s 52μT magnetic field.What is the magnitude of the emf induced between the two sides of the shark&#039;s head?Express your answer using two significant figures.

Matthew Rodriguez · Accepted Answer

Step 1Given:The velocity of hammerhead shark (v)=1.0msLength of hammerhead (L)=83cm.the magnetic field of the earth (B)=52μTthe angle between the hammerhead and the earth&#039;s magnetic field =90∘.Step 2the magnitude of electromotive force is given by,Femf=BL×vFemf=BLvsin⁡θFemf=52×10−6×83×10−2×1Femf=4.316×10−5NStep 3Answer:electromotive force =4.316×10−5N

Philip Williams · Accepted Answer

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=53×10−6T.  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=△la△t  Here △ is time interval.  The expression for the magnetic flux is,  △ϕa=Ba△Aa  The expression for the induced emf by using the Faradays

alenahelenash · Accepted Answer

Answer: 7.4×10−5T. 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=56×10−6T L=88×10−2m v=1.5m/s emf induced =56×10−6×88×10−2×1.5 =7.4×10−5T.

A scalloped hammerhead shark swims at a steady speed of

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