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Lisa [10]
3 years ago
5

A long iron bar lies along the x-axis and has current of I = 16.4 A running through it in the +x-direction. The bar is in the pr

esence of a uniform magnetic field, perpendicular to the current. There is a magnetic force per unit length on the bar of 0.132 N/m in the −y-direction. (a) What is the magnitude of the magnetic field (in mT) in the region through which the current passes?
Physics
1 answer:
Gre4nikov [31]3 years ago
8 0

Answer:

B = 8.0487mT

Explanation:

To solve the exercise it is necessary to take into account the considerations of the Magnetic Force described by Faraday,

The magnetic force is given by the formula

F =BILsin\theta

Where,

B = Magnetic Field

I = Current

L = Length

\theta = Angle between the magnetic field and the velocity, for this case are perpendicular, then is 90 degrees

According to our data we have that

I = 16.4A

F = 0.132N/m

As we know our equation must be modificated to Force per length unit, that is

\frac{F}{L} = BI sin(90)

Replacing the values we have that

0.132 = 16.4 (1) B

Solving for B,

B = \frac{0.132}{16.4}

B = 8.0487mT

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1.15 m/s to the left (3 sig. fig.).

<h3>Explanation</h3>

Momentum is conserved between the two balls if they are not in contact with any other object. In other words,

p_{\text{A,initial}} + p_{\text{B,initial}}=p_{\text{A,final}} + p_{\text{B,final}}

m_\text{A} \cdot v_{\text{A,initial}} + m_\text{B}\cdot v_{\text{B,initial}}=m_\text{A}\cdot v_{\text{A,final}} + m_\text{B}\cdot v_{\text{B,final}}, where

  • m stands for mass and
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Let the velocity of objects moving to the right be positive.

  • m_\text{A} = 1.55\;\text{kg},
  • m_\text{B} = 0.752\;\text{kg}.

Before the two balls collide:

  • v_\text{A} = +8.76\;\text{m}\cdot\text{s}^{-1},
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After the two balls collide:

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m_\text{A} \cdot v_{\text{A,initial}} + m_\text{B}\cdot v_{\text{B,initial}}=m_\text{A}\cdot v_{\text{A,final}} + m_\text{B}\cdot v_{\text{B,final}},

1.55 \times (+8.76) + 0.752 \times (-11.4) = 1.55\;{\bf v_{\textbf{A,final}}} + 0.752 \times (+9.03).

v_{\text{A,final}} = \dfrac{1.55 \times (+8.76) + 0.752 \times (-11.4)-0.752 \times (+9.03)}{1.55} = -1.15\;\text{m}\cdot\text{s}^{-1}.

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Hence, ball A will be travelling to the left at 1.15 m/s (3 sig. fig. as in the question) after the collision.

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