A 102-turn square coil of side 20.0 cm rotates about a vertical axis at = 1400 r

Question

A 102-turn square coil of side 20.0 cm rotates about a vertical axis at = 1400 rev/min as indicated in the figure below. The horizontal component of the Earth's magnetic field at the coil's location is equal to 2.00 105 T.

(a) Calculate the maximum emf induced in the coil by this field.
________ mV

(b) What is the orientation of the coil with respect to the magnetic field when the maximum emf occurs?

CHOICE A)The plane of the coil is parallel to the magnetic field.

CHOICE B)The plane of the coil is oriented 45° with respect to the magnetic field.     

CHOICE C)The plane of the coil is perpendicular to the magnetic field.

Explanation / Answer

Converting from rev/m to radians/s ..** = 1400^3/60(rev/s) x 2(rad/rev) =8796.459/60=146.60 radians/s

Assume coils each have an area A and a side-conductor length L.
Max emf is generated as L moves perpendicular to the field at a velocity v = r** (m/s)

The changing area swept out by a side conductor .. dA/dt (m²/s) = L(m) x v(m/s)
dA/dt = Lv = Lr ..
as Lr = ½ coil area A .. ..dA/dt = ½A per side conductor ..
as each coil has 2 side conductors .. .. dA/dt = A

From Faraday's law .. E = n B.dA/dt = nBA

E(max) = 102(2.0^-5T)(400^-4 m²)(146.60 radians/s).. E =14.073^-3 V (14.0mV)

(b)
Max emf is generated as L moves perpendicular to the field .. ie, as the plane of the coil is parallel with the field ..
As the field is horizontal, then dA/dt is a maximum as the plane of the coils becomes horizontal (remember it's not the max perp area that matters .. as when the coil plane is perp to field .. it's where the perp area is changing the fastest rate)

calculation may be wrong but concept is right

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