 David Troyer

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 $52\mu T$ magnetic field.
What is the magnitude of the emf induced between the two sides of the shark's head? Matthew Rodriguez

Expert

Step 1
Given:
The velocity of hammerhead shark $\left(v\right)=1.0\frac{m}{s}$
Length of hammerhead $\left(L\right)=83cm$.
the magnetic field of the earth $\left(B\right)=52\mu T$
the angle between the hammerhead and the earth's magnetic field $={90}^{\circ }$.
Step 2
the magnitude of electromotive force is given by,
${F}_{emf}=BL×v$
${F}_{emf}=BLv\mathrm{sin}\theta$
${F}_{emf}=52×{10}^{-6}×83×{10}^{-2}×1$
${F}_{emf}=4.316×{10}^{-5}N$
Step 3
electromotive force $=4.316×{10}^{-5}N$ Philip Williams

Expert

Given:
- The speed of scalloped hammerhead shark is: ${v}_{a}=1.0\frac{m}{s}$.
- The width of head is: ${w}_{a}=83cm=0.83m$.
- The magnetic field is: ${B}_{a}=52\mu T=53×{10}^{-6}T$.
The expression for the change in area through which magnetic field moves is,
$\mathrm{△}{A}_{a}=\mathrm{△}{l}_{a}{w}_{a}$
Here are change in length and width respectively.
The expression for the speed of scalloped hammerhead shark is,
${v}_{a}=\frac{\mathrm{△}{l}_{a}}{\mathrm{△}t}$
Here $\mathrm{△}$ is time interval.
The expression for the magnetic flux is,
$\mathrm{△}{\varphi }_{a}={B}_{a}\mathrm{△}{A}_{a}$
The expression for the induced emf by using the Faradays alenahelenash

Expert

Answer: $7.4×{10}^{-5}T$. 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 $B=56×{10}^{-6}T$ $L=88×{10}^{-2}m$ $v=1.5m/s$ emf induced $=56×{10}^{-6}×88×{10}^{-2}×1.5$ $=7.4×{10}^{-5}T$.