Explanation:
The emf is equal to the work done on the charge per unit charge (ϵ=dWdq) when there is no current flowing. Since the unit for work is the joule and the unit for charge is the coulomb, the unit for emf is the volt (1V=1J/C).
The magnitude of the magnetic field inside the solenoid is 3.4×10^(-4) T.
To find the answer, we need to know about the magnetic field inside the solenoid.
<h3>What's the expression of magnetic field inside a solenoid?</h3>
- Mathematically, the expression of magnetic field inside the solenoid= μ₀×n×I
- n = no. of turns per unit length and I = current through the solenoid
<h3>What's is the magnetic field inside the solenoid here?</h3>
- Here, n = 290/32cm or 290/0.32 = 906
I= 0.3 A
- So, Magnetic field= 4π×10^(-7)×906×0.3 = 3.4×10^(-4) T.
Thus, we can conclude that the magnitude of the magnetic field inside the solenoid is 3.4×10^(-4) T.
Learn more about the magnetic field inside the solenoid here:
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Answer:
at resonance impedence is equal to resistance and quality factor is dependent on R L AND C all
Explanation:
we know that for series RLC circuit impedance is given by
but we know that at resonance
putting in impedance formula , impedance will become
Z=R so at resonance impedance of series RLC is equal to resistance only
now quality factor of series resonance is given by
so from given expression it is clear that quality factor depends on R L and C
Answer:
The right solution is:
(a) 2.87 eV
(b) 1.4375 eV
Explanation:
Given:
Wavelength,
= 433 nm
Potential difference,
= 1.43 V
Now,
(a)
The energy of photon will be:
E =
=
or,
=
=
(b)
As we know,
⇒
By substituting the values, we get
⇒
⇒
or,
⇒
⇒