Answer:
The magnitude of the magnetic torque on the coil is 1.98 A.m²
Explanation:
Magnitude of magnetic torque in a flat circular coil is given as;
τ = NIASinθ
where;
N is the number of turns of the coil
I is the current in the coil
A is the area of the coil
θ is the angle of inclination of the coil and magnetic field
Given'
Number of turns, N = 200
Current, I = 7.0 A
Angle of inclination, θ = 30°
Diameter, d = 6 cm = 0.06 m
A = πd²/4 = π(0.06)²/4 = 0.002828 m²
τ = NIASinθ
τ = 200 x 7 x 0.002828 x Sin30
τ = 1.98 A.m²
Therefore, the magnitude of the magnetic torque on the coil is 1.98 A.m²
Answer:
(b) The electrons, because they have the smaller momentum and, hence, the larger de Broglie wavelength
Explanation:
de Broglie wavelength λ = h / m v
Since both electrons and protons have same velocity , momentum mv will be less for electrons because mass of electron is less .
for electron , momentum is less so . Therefore de Broglie wavelength λ will be more for electrons .
Amount of diffraction that is angle of diffraction is proportional to λ
Therefore electrons having greater de Broglie wavelength will show greater diffraction.
Answer:
B
Explanation:
OOf we are doing this stuff atm
So if its faster at the front and slow at the back you can tell that its not slowing down because less of a force is there however at the front there is more of a force. Friction is low which means that its not makimg much contact so no sudden change of forces thats also why its B
