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:
Option B
Explanation:
Magnification of Microscope is
Mo= Magnification of objective lens and
Me= magnification of the eyepiece.
Both magnifications( of objective and eyepiece) are inversely proportional to the focal length.
Magnification,

when the focal length is less magnification will be high and when the magnification is the low focal length of the microscope will be more.
Thus. Magnification will increase by decreasing the focal length.
The correct answer is Option B
I agree with the other comment
Answer:
The velocity of the photo electron is
.
Explanation:
Given that,
Supplied energy, 
Minimum energy of the electron to escape from the metal, 
We need to find the velocity of the photo electron. The energy supplied by the photon is equal to the sum of minimum escape energy and the kinetic energy of the escaping electron. So,

The formula of kinetic energy is given by :

So, the velocity of the photo electron is
.
Answer:
80 J
Explanation:
Ep = mgh
Ep = (4 kg) (10 m/s²) (2 m)
Ep = 80 J