Question

A conducting wire formed in the shape of a right triangle with base b = 28 cm and height h = 83 cm and having resistance R = 2.8 ?, rotates uniformly around the y-axis in the direction indicated by the arrow (clockwise as viewed from above (loooking down in the negative y-direction)). The triangle makes one complete rotaion in time t = T = 1.2 seconds. A constant magnetic field B = 1.1 T pointing in the positive z-direction (out of the screen) exists in the region where the wire is rotating.

1)
What is ?, the angular frequency of rotation?

2)
What is Imax, the magnitude of the maximum induced current in the loop?

3)
At time t = 0, the wire is positioned as shown. What is the magnitude of the magnetic flux ?1 at time t = t1 = 0.45 s?

4)
What is I1, the induced current in the loop at time t = 0.45 s? I1 is defined to be positive if it flows in the negative y-direction in the segment of length h.

Answer 1

Note: You didn't provide the diagram referred to but i can help you work around it without the diagram.

Answer:

1) Angular frequency of rotation = 5.24 rad/s

2)
`I_(max) = 0.24 A`

3) Magnitude of the magnetic flux = 0.091 wb

4) The induced current in the loop at t = 0.45, I₁ = 0.169 A

Step-by-step explanation:

1) Angular frequency of rotation, w, is given by the formula:

`\omega = (2 \pi)/(T)`

Period for one complete rotation, T = 1.2 seconds

`\omega = (2 \pi)/(1.2) \\\omega = 5.24 rad/s`

b) The magnitude of the maximum current induced in the loop:

I = e/R

Where e = induced emf, and Resistance, R = 2.8 cm

e = NBAw

Since there is one complete rotation, N = 1

Magnetic field, B = 1.1 T

base, b = 28 cm = 0.28 m

height, h = 83 cm = 0.83 m

Area of the triangle, A = 0.5 * base * height

A = 0.5 * 0.28 * 0.83

A = 0.1162 m²

e = 1 * 1.1 * 0.1162 * 5.24

e = 0.6698 V

`I_(max) = (e sin 90)/(R) \\I_(max) = (0.6698 * 1)/(2.8)\\I_(max) = 0.24 A`

3)

θ is the angle between the normal and the magnetic field

At t = 0, θ = 0

At t = 0.45 s, θ₁ = ?

`\omega = (\theta_(1) - \theta )/(t_(1) -t)\\5.24 = (\theta_(1) - 0 )/(0.45 -0)\\5.24 = (\theta_(1) )/(0.45)\\\theta_1 = 5.24 * 0.45\\\theta_1 = 2.358 rad\\\theta_1 = 2.358 * 180/\pi \\\theta_1 = 135.1^(0)`

The magnetic flux is calculated as:

`\phi = NBA cos \theta_(1) \\\phi = 1*1.1 * 0.1162 cos135.1\\\phi = -0.091 wb`

Magnitude of the magnetic flux = 0.091 wb

4)

Induced current, I = e/R

e = dΦ/dt

Φ = NBA cos(wt)

e = dΦ/dt = -NBAw sin (wt)

e = -(1 * 1.1 * 0.1162*5.24) sin(5.24 * 0.45*180/π)

e = -0.4736 V

I₁ = -0.4736/2.8

I₁ = -0.169 A

I₁ is defined to be positive due to the direction of flow

I₁ = 0.169 A