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3 years ago

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At a particular instant, a proton at the origin has velocity < 5e4, -2e4, 0> m/s. You need to calculate the magnetic field

at location < 0.03, 0.05, 0 > m, due to the moving proton. What is the vector r?

1 answer:

vesna_86 [32]3 years ago

3 0

Answer:

$9.7\times 10^{-5} T$

Explanation:

Velocity = $5\times 10^4i-2\times 10^4j$

r= $0.03i+0.05j$

r= $\mid r\mid=\sqrt{(0.03)^2+(0.05)^2}=0.058$

v= $\mid V\mid=\sqrt{(5\times 10^4)^2+(-2\times 10^{4})^2}=5.39\times 10^{2}$

We know that

$B=\frac{mv}{qr}$

Where q= $1.6\times 10^{-19} C$

Mass of proton= $1.67\times 10^{-27} kg$

Using the formula

$B=\frac{1.67\times 10^{-27}\times 5.39\times 10^2}{1.6\times 10^{-19}\times 0.058}$

$B=9.7\times 10^{-5} T$

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Answer:

The pickup truck and hatchback will meet again at 440.896 m

Explanation:

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Hatchback:

$a=5m/s^{2}$ , $v_{i}$ = 0 m/s (since initial velocity is zero), $s_{o}$ = 0

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putting in the data we will get

$s=(1/2)(5)t^{2}$

now putting 's' value from eq (1)

$2.5t^{2}-33.2t=0$

which will give,

t = 13.28 s

so both vehicles will meet up gain after 13.28 sec.

putting t = 13.28 in eq (1) will give

s = 440.896 m

So, both vehicles will meet up again at 440.896 m.

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