The emf is induced in the wire will be 1.56 ×10 ⁻³ V. The induced emf is the product of the magnetic field,velocity and length of the wire.
<h3>What is induced emf?</h3>
Emf is the production of a potential difference in a coil as a result of changes in the magnetic flux passing through it.
When the flux coupling with a conductor or coil changes, electromotive Force, or EMF, is said to be induced.
The given data in the problem is;
B is the magnitude of the magnetic field,= 5.0 ×10⁻⁵ T
V(velocity)=125 M/SEC
L(length)=25 cm=0.25 m
The maximum emf is found as;
E=VBLsin90°
E=125 × 5.0 × 10⁻⁵ ×0.25
E=1.56 ×10 ⁻³ V
Hence, the emf is induced in the wire will be 1.56 ×10 ⁻³ V
To learn more about the induced emf, refer to the link;
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Is there any types of answer to get an idea
Answer:
k = 2.279
Explanation:
Given:
Magnitude of charge on each plate, Q = 172 μC
Now,
the capacitance, C of a capacitor is given as:
C = Q/V
where,
V is the potential difference
Thus, the capacitance due to the charge of 172 μC will be
C = 
Now, when the when the additional charge is accumulated
the capacitance (C') will be
C' = 
or
C' = 
now the dielectric constant (k) is given as:
substituting the values, we get
or
k = 2.279
The relationship between current and voltage and resistance is described by ohlm's law. This equation i=v/r tells that the current i flowing through a circuit is directly proportional to the voltage v, and inversely proportional to resistance r. This desceibes the relationship of voltage, current and resistance.
