Option (B) is correct.The magnetic field strength=8.5 x 10² T
Explanation:
the magnetic force Fm is given by
Fm= q V B sinθ
q= charge=1.4 x 10⁻⁷ C
v= velocity= 1.3 x 10⁶ m/s
B= magnetic field strength
Fm= magnetic force= 1.5 x 10² N
θ=75°
so 1.5 x 10²=(1.4 x 10⁻⁷) (1.3 x 10⁶ ) (B) sin75
B=8.5 x 10² T
<h3>Given, </h3>
Force,F = 4000 N
Area,a = 50 m²
<h3>We know that, </h3>
Pressure = Force/Area
★ Putting the values in the above formula,we get:


Answer:
the frequency of the second harmonic of the pipe is 425 Hz
Explanation:
Given;
length of the open pipe, L = 0.8 m
velocity of sound, v = 340 m/s
The wavelength of the second harmonic is calculated as follows;
L = A ---> N + N--->N + N--->A
where;
L is the length of the pipe in the second harmonic
A represents antinode of the wave
N represents the node of the wave

The frequency is calculated as follows;

Therefore, the frequency of the second harmonic of the pipe is 425 Hz.
Answer:
Voltage in primary coil is 3.91 V
Explanation:
For transformer we know that the working principle is given as

here we know that
![V_1 [tex] = voltage in primary coil[tex]V_2 = 25 V](https://tex.z-dn.net/?f=V_1%20%5Btex%5D%20%3D%20voltage%20in%20primary%20coil%3C%2Fp%3E%3Cp%3E%5Btex%5DV_2%20%3D%2025%20V)


Now we have


To solve this problem we will apply the concept related to the magnetic dipole moment that is defined as the product between the current and the object area. In our case we have the radius so we will get the area, which would be



Once the area is obtained, it is possible to calculate the magnetic dipole moment considering the previously given definition:



Therefore the magnetic dipole moment is 