4. Layman is taking a day off from work and go to the beach, where he meets up w

Question

4. Layman is taking a day off from work and go to the beach, where he meets up with Feynman. After attending Feynman's recent lecture on "magnetic force on charge particles", Layman jokingly ask Feynman if he can apply that concept at the beach. In response, Feynman informs Layman that the saltiness of the ocean is due to dissolved ions, and that they in fact experience a magnetic force due to the presence of the Earth's magnetic eld. As a consequence of this force, negative and positive ions move in opposite directions until the electric field due to the charge separation balances the magnetic force. Furthermore, the charge separation results in a voltage that is measurable and pure excitement for the physicists. While picking up their surf boards, Feynman informs Layman that the California Current moves southward along the californian coastline with a speed of 0.25 m/s (in comparison, the Gulf Stream along the atlantic coast moves at ~ 3.5 m/s). The Earth's nnagnetic field at the specific location has a strength of 50.0 T and directed at an angle of 60.0° below the horizontal along the meridian. (a) What is the direction of the magnetic force on a single ionized negative chlorine ion moving in the California Current? (b) What is the magnitude of this magnetic force? (c) What electric field is required to balance the magnetic force? (d) If you place an electrode 5 m below the surface, where should you place a second electrode such that you lirection). Explain your answer.

Explanation / Answer

(a)The direction of the force can be found out by finding the direction of (v x b). As Fb=q*(v x b)

which would be perpendicularly upwards.

(b) As depicted above

F=q*(v x b) =q*v*b*sin(theta) {q is charge, v is velocity of particle, b is the strength of magnetic field}

here theta = 60

F=1.6*10^-19 * .25* 50*10^-6 * sin(60) {since single ionized chlorine ion as charge equal to e=1.6*10^-19}

F=1.732*10^(-24) N

(c) force due to electric field Fe= q*E {E is the strength of elctric field}

Fe=Fb

q*E=q*v*b*sin(theta)

E=v*b*sin(theta)

E= 0.25*50*10^-6*sin(60)

E=1.08*10^(-5) Newton/coulomb.

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