Q35E A long wire carrying 4.50 A of c... [FREE SOLUTION]

91影视

University Physics with Modern Physics

Hugh D. Young, Roger A. Freedman

Physics

14th edition Edition 路 1596 pages

ISBN: 9780321973610

Chapter 4: Q35E (page 914)

A long wire carrying 4.50 A of current makes two 90掳 bends, as shown in Fig. E27.35. The bent part of the wire passes through a uniform 0.240-T magnetic field directed as shown in the figure and confined to a limited region of space. Find the magnitude and direction of the force that the magnetic field exerts on the wire.

Short Answer

Expert verified

The magnitude of a long wire carrying current of 4.5 A and making a bend of $90 掳$

is 0.724N and the angle is $63.4 掳$

Step by step solution

Horizontal force

The force acting on the first and the third segment are all in the same direction that is downwards acting on the second segment which points to the right direction, Between the magnetic field and the length is , now using:

$F = l l B \sin 蠒$

we can write the vertical forces as:

$\begin{aligned} F_{y} = {IL}_{a} B + {IL}_{c} B \\ = (4 .50) 脳 (0 .240) + (4 .50 A) (60 .0 cm 鈭� x) (0 .240 T) \\ = (4 .50 A) (0 .60 mm) (0 .240 T) \\ = 0 .648 N \\ Now, the horizontal force is : \\ F_{x} = ILB \\ = (4 .50 A) (0 .30 m) (0 .240 T) \\ = 0 .324 T \end{aligned}$

Now, the horizontal force is:

Calculating the force and the and the angle

$\begin{aligned} \tan 鈦� 胃 = \frac{F_{y}}{F_{x}} \\ = \frac{0 .648}{0 .324} \\ = 2 .00 \end{aligned}$

Therefore, the angle of the wire is:

$\begin{aligned} 胃 = 2.00 \\ = {63.4}^{鈭�} \end{aligned}$

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Short Answer

Step by step solution

Horizontal force

Calculating the force and the and the angle

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