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adoni [48]
3 years ago
15

At a certain location, Earth has a magnetic field of 0.60 ✕ 10−4 T, pointing 75° below the horizontal in a north-south plane. A

15.0 m long straight wire carries a 19 A current. (a) If the current is directed horizontally toward the east, what are the magnitude and direction of the magnetic force on the wire? magnitude N
Physics
1 answer:
saveliy_v [14]3 years ago
5 0

Answer with Explanation:

We are given that

Magnetic field,B=0.6\times 10^{-4} T

\theta=75^{\circ}

Length of wire,l=15 m

Current,I=19 A

a.We have to find the magnitude of magnetic force and direction of magnetic force.

Magnetic force,F=IBlsin\theta

Using the formula

F=0.6\times 10^{-4}\times 15\times 19sin75

F=16.5\times 10^{-3} N

Direction=tan\theta=cot(90-75)=tan15^{\circ}

\theta=15^{\circ}

15 degree above the horizontal  in the northward direction.

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Asumamos que el cambio dimensional como consecuencia de la temperatura es pequeña, entonces podemos estimar el área de la placa de cobre en función de la temperatura mediante la siguiente aproximación:

A_{f} = w\cdot l \cdot [1 + 2\cdot \alpha\cdot (T_{f}-T_{o})] (1)

Donde:

w - Ancho de la placa, en centímetros.

l - Longitud de la placa, en centímetros.

\alpha - Coeficiente de dilatación, en \frac{1}{^{\circ}C}.

T_{o} - Temperatura inicial, en grados Celsius.

T_{f} - Temperatura final, en grados Celsius.

Si sabemos que w = 65\,cm, l = 78\,cm, \alpha = 17\times 10^{-6}\,\frac{1}{^{\circ}C}, T_{o} = 20\,^{\circ}C and T_{f} = 400\,^{\circ}C, entonces el área de la placa a la temperatura final:

A_{f} = (65\,cm)\cdot (78\,cm)\cdot \left[1+\left(17\times 10^{-6}\,\frac{1}{^{\circ}C} \right)\cdot (400\,^{\circ}C-20\,^{\circ}C)\right]

A_{f} = 5102.752\,cm^{2}

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