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3 years ago

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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?

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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"A circular coil of radius r = 5 cm and resistance R = 0.2 ? is placed in a uniform magnetic field perpendicular to the plane of the coil. The magnitude of the field changes with time according to B = 0.5 e^-t T. What is the magnitude of the current induced in the coil at the time t = 2 s?"

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we need to determine the emf induced in the coil and y applying ohm's law we determine the current induced.

using the formula be low,

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