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3 years ago

14

An eight-turn coil encloses an elliptical area having a major axis of 40.0 cm and a minor axis of 30.0 cm. The coil lies in the

plane of the page and carries a clockwise current of 6.20 A. If the coil is in a uniform magnetic field of 1.98 10-4 T directed toward the left of the page, what is the magnitude of the torque on the coil? Hint: The area of an ellipse is A = ?ab, where a and b are, respectively, the semimajor and semiminor axes of the ellipse.

1 answer:

Darina [25.2K]3 years ago

5 0

Answer:

9.25 x 10^-4 Nm

Explanation:

number of turns, N = 8

major axis = 40 cm

semi major axis, a = 20 cm = 0.2 m

minor axis = 30 cm

semi minor axis, b = 15 cm = 0.15 m

current, i = 6.2 A

Magnetic field, B = 1.98 x 10^-4 T

Angle between the normal and the magnetic field is 90°.

Torque is given by

τ = N i A B SinФ

Where, A be the area of the coil.

Area of ellipse, A = π ab = 3.14 x 0.20 x 0.15 = 0.0942 m²

τ = 8 x 6.20 x 0.0942 x 1.98 x 10^-4 x Sin 90°

τ = 9.25 x 10^-4 Nm

thus, the torque is 9.25 x 10^-4 Nm.

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Anon25 [30]

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Explanation:

Let first consider all data that are given in question.

1. F = 8 N ...force acting on string

2. f = 2 Hz ...frequency of system

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Under certain conditions, waves can bounce back and forth through a particular region, effectively becoming stationary. These are called standing waves.

Here,it is due to vibration induced in spring due to tension induced in string

Standing wave equation is given by

$y = (x,t) = 2A * sinK x * cos (wt)$ ...(1)

Let first find, value of K, x, w, t

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Where β is wavelength in meters

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now, let us find value of w

W = 2 x pi x f ....(3)

where f is frequency in hertz

W = 2 x pi x 2

W = 4 x pi

W = 0.08 $\frac{m}{s}$

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now, let us find value of v that is wave speed

Notice that some x-positions of the resultant wave are always zero no matter what the phase relationship is. These positions are called nodes.

Finding the positions where the sine function equals zero provides the positions of the nodes.

In our case, and

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