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Otrada [13]
2 years ago
8

The strength of the Earth’s magnetic field B at the equator is approximately equal to 5 × 10−5 T. The force on a charge q moving

in a direction perpendicular to a magnetic field is given by F = q v B, where v is the speed of the particle. The direction of the force is given by the right-hand rule. Suppose you rub a balloon in your hair and your head acquires a static charge of 6 × 10−9 C. If you are at the equator and driving west at a speed of 80 m/s, what is the strength of the magnetic force on your head due to the Earth’s magnetic field? Answer in units of N.
Physics
1 answer:
Blizzard [7]2 years ago
7 0

Answer:

2.4\cdot 10^{-11} N

Explanation:

Since the Earth's magnetic field is perpendicular to your direction of motion, the strength of the magnetic force exerted on your head is given by:

F=qvB

where:

q=6\cdot 10^{-9}C is the charge on your head

v=80 m/s is the speed at which you are moving

B=5\cdot 10^{-5} T is the strength of the magnetic field of the Earth

By substituting these numbers into the equation, we find the strength of the magnetic force:

F=(6\cdot 10^{-9}C)(80 m/s)(5\cdot 10^{-5} T)=2.4\cdot 10^{-11} N

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2 years ago
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pantera1 [17]

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\alpha=\cos^{-1}(0.3846)

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50^2=130^2+120^2-2\times130\times120\cos\beta

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\beta=\cos^{-1}(0.923)

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We need to calculate the force on 130 cm side

Using formula of force

F_{130}=ILB\sin\theta

F_{130}=4\times130\times10^{-2}\times75\times10^{-3}\sin0

F_{130}=0

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Using formula of force

F_{120}=ILB\sin\beta

F_{120}=4\times120\times10^{-2}\times75\times10^{-3}\sin22.63

F_{120}=0.1385\ N

The direction of force is out of page.

We need to calculate the force on 50 cm side

Using formula of force

F_{50}=ILB\sin\alpha

F_{50}=4\times50\times10^{-2}\times75\times10^{-3}\sin67.38

F_{50}=0.1385\ N

The direction of force is into page.

Hence, The magnitude of the magnetic force on each of the three sides of the loop are 0 N, 0.1385 N and 0.1385 N.

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2 years ago
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