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3 years ago

What is the unit for the volt? W V = 9 What is the unit for the volt

Physics

1 answer:

Likurg_2 [28]3 years ago

4 0

the unit for volt is v

The volt (symbol: V) is the derived unit for electric potential, electric potential difference (voltage), and electromotive force.

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A person drives 70 km/h in 1 hour to the east, then 80 km/h for another hour to the east. What

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3 years ago

Read 2 more answers

A proton and an electron are moving in the +x direction in a magnetic field in the +z

Aleksandr [31]

Answer:

Explanation:

Force on a moving charge is given by the following relation

F = q ( v x B )

for proton

q = e , v = vi , B = Bk

F = e ( vi x Bk )

= Bev - j

= - Bevj

The direction of force is along negative of y axis or -y - axis.

for electron

q = - e , v = vi , B = Bk

F = - e ( vi x Bk )

= - Bev - j

= Bevj

The direction of force is along positive of y axis or + y - axis.

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3 years ago

A long, straight, horizontal wire carries a left-to-right current of 40 A. If the wire is placed in a uniform magnetic field of

Drupady [299]

Answer:

$4.5\times 10^{-5} T$

Explanation:

We are given that

Current in wire=40 A

Magnetic field= $B_1=3.5\times 10^{-5}$ T( vertically downward)

We have to find the resultant magnitude of the magnetic field 29 cm above the wire and 29 cm below the wire.

According to Bio-Savart law, the magnetic field exerted by the wire at distance R is given by

$B_{wire}=B_2=\frac{\mu_0I}{2\pi R}$

We have R=29 cm= $\frac{29}{100}=0.29 m$

1 m=100 cm

Substitute the values in the given formula

$B_2=\frac{4\pi\times 10^{-7}\times 40}{2\times \pi\times 0.29}=\frac{2\times 40\times 10^{-7}}{0.29}=2.76\times 10^{-5} T$

The resultant magnetic field is given by

$B=\sqrt{B^2_1+B^2_2}$

Substitute the values then we get

$B=\sqrt{(3.5\times 10^{-5})^2+(2.76\times 10^{-5})^2}$

$B=4.5\times 10^{-5} T$

The resultant magnitude of magnetic field is same above and below the wire as it is at same distance.

The resultant magnitude of the magnetic field 29 cm below the wire= $4.5\times 10^{-5} T$

Hence, the resultant magnitude of the magnetic field 29 cm above the wire= $4.5\times 10^{-5} T$

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3 years ago