Question

A 65-cm segment of conducting wire carries a current of 0.35

Answer 1

The intensity of the magnetic force exerted on the wire due to the presence of the magnetic field is given by
$F=ILB \sin \theta$
where
I is the current in the wire
L is the length of the wire
B is the magnetic field intensity
$\theta$ is the angle between the direction of the wire and the magnetic field

In our problem, L=65 cm=0.65 m, I=0.35 A and B=1.24 T. The force on the wire is F=0.26 N, therefore we can rearrange the equation to find the sine of the angle:
$\sin \theta= \frac{F}{ILB}= \frac{0.26 N}{(0.35 A)(0.65 m)(1.24 T)}=0.922$

and so, the angle is
$\theta=\arcsin(0.921)=67.1^{\circ}$

Answer 2

The angle between the wire segment and the magnetic field is about 67°

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Further explanation

Let's recall magnetic force on conducting wire as follows:

$\boxed{F = B I L \sin \theta}$

where:

F = magnetic force ( N )

B = magnetic field strength ( T )

I = current ( A )

L = length of wire ( m )

θ = angle between magnetic field and the wire segment

Let's tackle the problem now !

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Given:

length of wire = L = 65 cm = 0.65 m

current = I = 0.35 A

magnetic field strength = B = 1.24 T

magnitude of force = F = 0.26 N

Asked:

angle between magnetic field and the wire segment = θ = ?

Solution:

$F = B I L \sin \theta$

$0.26 = 1.24 \times 0.35 \times 0.65 \times \sin \theta$

$\sin \theta = \frac{200}{217}$

$\theta = \sin^{-1} (\frac{200}{217})$

$\boxed{\theta \approx 67^o}$

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Conclusion:

The angle between the wire segment and the magnetic field is about 67°

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Learn more

• Temporary and Permanent Magnet : brainly.com/question/9966993
• The three resistors : brainly.com/question/9503202
• A series circuit : brainly.com/question/1518810
• Compare and contrast a series and parallel circuit : brainly.com/question/539204

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Answer details

Grade: High School

Subject: Physics

Chapter: Magnetic Field