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hodyreva [135]
4 years ago
8

The ocean is salty because it contains many dissolved ions. As these charged particles move with the water in strong ocean curre

nts, they feel a force from the earthâs magnetic field. Positive and negative charges are separated until an electric field develops that balances this magnetic force. This field produces measurable potential differences that can be monitored by ocean researchers. The Gulf Stream moves northward off the east coast of the United States at a speed of up to 3.5 m/s . Assume that the current flows at this maximum speed and that the earthâs field is 50 μT tipped 60 degrees below horizontal.
Part A
What is the direction of the magnetic force on a singly ionized negative chlorine ion moving in this ocean current? Choose best answer
(a) east
(b) west
(c) up
(d) down
Part BWhat magnitude electric field is necessary to exactly balance this magnetic force?
Choose best answer.
(a) 1.8 x 10-4 N/C
(b) 1.5 x 10-4 N/C
(c) 1.0 x 10-4 N/C
Physics
1 answer:
Yuliya22 [10]4 years ago
8 0

Given that,

Speed = 3.5 m/s

Magnetic field = 50μT

Angle = 60°

(A). We need to find the direction of magnetic force

Using formula of magnetic force

\vec{F}=q(\vec{v}\times\vec{B})

Here, (\vec{v}\times\vec{B})= down

But , charge is negative.

So, the direction of magnetic force will be up.

(B). We need to calculate the magnetic electric field

Using formula of magnetic force

F=qvB\sin\theta

qE=qvB\sin\theta

E=vB\sin\theta

Where, v = speed

B = magnetic field

Put the value into the formula

E=3.5\times50\times10^{-6}\sin60

E=0.000151\ N/C

E=1.5\times10^{-4}\ N/C

Hence, (A). The direction of magnetic force is UP

(c) is correct option

(B). The magnetic electric field is 1.5\times10^{-4}\ N/C

(b) is correct option

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Answer:

v_{0}=319.2 m/s    

Explanation:

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Conservation of energy.

We have kinetic energy at first and kinetic and potential energy at the end.            

(1/2)(m+M)V_{1}^{2}=(1/2)(m+M)V_{2}^{2}+(m+M)gh

Here:

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We can find V(2) using the definition of force at this point:

\Sigma F=(m+M)a_{c}=(m+M)(V_{2}^{2}/R)

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Now, we can solve the equation to find V(2)

V_{2}=\sqrt{\frac{R*(T-(m+M)*g*cos(\theta))}{(m+M)}}

V_{2}=\sqrt{\frac{1.72*(4.86-(0.01+0.75)*9.81*cos(57.66))}{(0.01+0.75)}}

V_{2}=1.40 m/s        

Now we can find V(1) using the conservation of energy equation

(1/2)V_{1}^{2}=(1/2)V_{2}^{2}+gh

V_{1}=\sqrt{V_{2}^{2}+2gh}

V_{1}=\sqrt{1.40^{2}+2*9.81*0.8}          

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v_{0}=\frac{(m+M)V_{1}}{m}                

v_{0}=\frac{(0.01+0.75)*4.20}{0.01}

v_{0}=319.2 m/s        

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Explanation:

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