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BigorU [14]
3 years ago
10

A vertical straight wire 35.0 cmcm in length carries a current. You do not know either the magnitude of the current or whether t

he current is moving upward or downward. If there is a uniform horizontal magnetic field of 0.0300 TT that points due north, the wire experiences a horizontal magnetic force to the west of 0.0180 NN. Find the magnitude of the current.
Physics
1 answer:
12345 [234]3 years ago
8 0

Answer:

1.714\ \text{A}

Explanation:

F = Magnetic force = 0.018 N

B = Magnetic field = 0.03 T

L = Length of wire = 35 cm

\theta  = Angle between current and magnetic field = 90^{\circ}

Magnetic force is given by

F=IBL\sin\theta\\\Rightarrow I=\dfrac{F}{BL\sin\theta}\\\Rightarrow I=\dfrac{0.018}{0.03\times 35\times 10^{-2}\times \sin90^{\circ}}\\\Rightarrow I=1.714\ \text{A}

The magnitude of the current is 1.714\ \text{A}.

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A source containing a mixture of hydrogen and deuterium atoms emits light at two wavelengths whose mean is 540 nm and whose sepa
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Answer:

N=3176.5rulling

Explanation:

We were told that the source containing a mixture of hydrogen and deuterium atoms emits light with

wavelengths whose mean is 540 nm

Then λ= 540 nm, but we need to convert to metre which = (540× 10⁻⁹m)

Also whose separation is 0.170 nm, which mean the difference between the wavelength is 0.170 nm then

Δ λ = 0.170 nm the we convert to metre we have. Δλ= 0.170 nm= (0.170×10⁻⁹m)

the formular below can be used to can be used to calculate our minimum number of lines

N= λ /(m Δλ)

Where N is number of fillings i.e number of lines

λ= wavelength

Δλ= difference in wavelength

m=1

Then if we substitute the values we have

,N= (540× 10⁻⁹ m)/[(1)*(0.170× 10⁻⁹m)]

N =3176.5rulling

Therefore, minimum number of lines = =3176.5rulling

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Coherent light with wavelength 598 nm passes through two very narrow slits, and the interference pattern is observed on a screen
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Answer:

1.196 μm

Explanation:

D = Screen distance = 3 m

\lambda = Wavelength = 598 m

y = Distance of first-order bright fringe from the center of the central bright fringe = 4.84 mm

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tan\theta=\frac{y}{D}\\\Rightarrow \theta=tan^{-1}{\frac{y}{D}}\\\Rightarrow \theta=tan^{-1}{\frac{4.84\times 10^{-3}}{3}}\\\Rightarrow \theta=0.09243\ ^{\circ}

sin\theta=\frac{\lambda}{d}\\\Rightarrow d=\frac{\lambda}{sin\theta}\\\Rightarrow d=\frac{598\times 10^{-9}}{sin0.09243}\\\Rightarrow d=0.00037066\ m

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dsin\theta=\frac{\lambda'}{2}\\\Rightarrow \lambda'=2dsin\theta\\\Rightarrow \lambda'=2\times 0.00037066\times sin0.09243\\\Rightarrow \lambda'=1.196\times 10^{-6}\\\Rightarrow \lambda'=1.196\ \mu m

Wavelength of first-order dark fringe observed at this same point on the screen is 1.196 μm

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