A stationary circular wall clock has a face with a radius of 12 cm. Six turns of wire are wound around its perimeter; the wire carries a current of 2.4 A in the clockwise direction. The clock is located where there is a constant, uniform external magnetic field of magnitude 82 mT (but the clock still keeps perfect time). At exactly 1:00 P.M., the hour hand of the clock points in the direction of the external magnetic field. (a) After how many minutes will the minute hand point in the direction of the torque on the winding due to the magnetic field? (b) Find the torque magnitude.

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aristocles aristocles · Answer 1 · 2019-11-03T03:32:32+01:00

Answer:

Part a)

torque is perpendicular to the magnetic field

so minute hand will be at 90 degree to hour hand so it will be after t = 20 minutes

Part b)

[tex]\tau = 0.0534 Nm[/tex]

Explanation:

Part a)

As we know that external magnetic field is along the hour hand when it shows time of 1:00 PM

so we know that

magnetic field makes 30 degree angle with vertical in the the plane of clock

Now current in the turns of the clock is in clockwise direction

so we have magnetic moment is perpendicular to the plane of the clock and inward in the direction

now we have

[tex]\tau = \vec M \times \vec B[/tex]

so torque is perpendicular to the magnetic field

so minute hand will be at 90 degree to hour hand so it will be after t = 20 minutes

Part b)

[tex]\tau = NiAB[/tex]

here we know that

N = 6

i = 2.4 A

[tex]A = \pi(0.12)^2[/tex]

B = 82 mT

so we have

[tex]\tau = 6(2.4)(\pi (0.12)^2)(82 \times 10^{-3})[/tex]

[tex]\tau = 0.0534 Nm[/tex]