Answered

Suppose a 56-turn coil lies in the plane of the page in a uniform magnetic field that is directed out of the page. The coil originally has an area of 0.275 m2. It is stretched to have no area in 0.100 s. What is the magnitude (in V) and direction (as seen from above) of the average induced emf if the uniform magnetic field has a strength of 1.00 T?

Answer :

Answer:

The average induced emf is 154 V .

Explanation:

Given that,

Number of turns = 56

Area = 0.275 m²

Time = 0.100 s

Magnetic field = 1.00 T

We need to calculate the magnetic flux

Using formula of magnetic flux

[tex]\phi=B\cdot\Delta A[/tex]

We need to calculate the emf

Using formula of emf

[tex]\epsilon_{0}=-N\dfrac{d\phi}{dt}[/tex]

[tex]\epsilon_{0}=NB\dfrac{\Delta A}{dt}[/tex]

[tex]\epsilon_{0}=56\times1.00\dfrac{0.275}{0.100}[/tex]

[tex]\epsilon_{0}=154\ V[/tex]

Since the magnetic field is directed into the plane of the paper.An opposite magnetic field is produced by the induced emf. So the current must be in the anticlockwise direction so as to oppose the applied magnetic field.

The direction of the current anticlockwise

Hence, The average induced emf is 154 V .

Other Questions