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Elan Coil [88]
2 years ago
10

A 25-turn circular coil of wire has diameter 1.00 m. It is placed with its axis along the direction of the Earth’s magnetic fiel

d of 50.0 μT and then in 0.200 s is flipped 180°. An average emf of what magnitude is generated in the coil?
Physics
1 answer:
slava [35]2 years ago
3 0

Answer:

The magnitude of average emf is 9.813 mV.

Explanation:

Given that,

Number of turns = 25 turns

Diameter = 1.00 m

Magnetic field = 50.0μm

Time = 0.200 s

Angle = 180°

We need to calculate the area of the coil

Using formula of area

A=\pr r^2

A=\pi\times(\dfrac{1.00}{2})^2

A=0.785\ m^2

We need to calculate the average emf

Using formula of emf

E=NAB\cos\theta

E=NA(\dfrac{B\cos0^{\circ}-B\cos180^{\circ}}{dt}

E=\dfrac{2NA}{dt}

Where, N = number of turns

A = Area

B = magnetic field

Put the value into the formula

E=\dfrac{2\times25\times0.785\times50.0\times10^{-6}}{0.200}

E=9.813\ mV

Hence, The magnitude of average emf is 9.813 mV.

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