Answer:
Explanation:
Force on a current carrying rod due to magnetic field is given as
here we know that
current in the rod
now magnetic force is balanced by the weight of the rod
so we will have
Answer:
the 70kg man
Explanation:
because he has more weight and is moving faster
Answer:
1.77 x 10^-8 C
Explanation:
Let the surface charge density of each of the plate is σ.
A = 4 x 4 = 16 cm^2 = 16 x 10^-4 m^2
d = 2 mm
E = 2.5 x 10^6 N/C
ε0 = 8.85 × 10-12 C2/N ∙ m2
Electric filed between the plates (two oppositively charged)
E = σ / ε0
σ = ε0 x E
σ = 8.85 x 10^-12 x 2.5 x 10^6 = 22.125 x 10^-6 C/m^2
The surface charge density of each plate is ± σ / 2
So, the surface charge density on each = ± 22.125 x 10^-6 / 2
= ± 11.0625 x 10^-6 C/m^2
Charge on each plate = Surface charge density on each plate x area of each plate
Charge on each plate = ± 11.0625 x 10^-6 x 16 x 10^-4 = ± 1.77 x 10^-8 C
Answer:
Explanation:
From the data it appears that A is the middle point between two charges.
First of all we shall calculate the field at point A .
Field due to charge -Q ( 6e⁻ ) at A
= 9 x 10⁹ x 6 x 1.6 x 10⁻¹⁹ / (2.5)² x 10⁻⁴
= 13.82 x 10⁻⁶ N/C
Its direction will be towards Q⁻
Same field will be produced by Q⁺ charge . The direction will be away
from Q⁺ towards Q⁻ .
We shall add the field to get the resultant field .
= 2 x 13.82 x 10⁻⁶
= 27.64 x 10⁻⁶ N/C
Force on electron put at A
= charge x field
= 1.6 x 10⁻¹⁹ x 27.64 x 10⁻⁶
= 44.22 x 10⁻²⁵ N
The average current density in the wire is given by:
where I is the current intensity and A is the cross-sectional area of the wire.
The cross-sectional area of the wire is given by:
where r is the radius of the wire. In this problem, 
, so the cross-sectional area is
and the average current density is
