Explanation:
For this problem we have to take into account the expression
J = I/area = I/(π*r^(2))
By taking I we have
I = π*r^(2)*J
(a)
For Ja = J0r/R the current is not constant in the wire. Hence

and on the surface the current is

(b)
For Jb = J0(1 - r/R)

and on the surface

(c)
Ja maximizes the current density near the wire's surface
Additional point
The total current in the wire is obtained by integrating

and in a simmilar way for Jb
![I_{T}=\pi J_{0} \int\limits^R_0 {r^{2}(1-r/R)} \, dr = \pi J_{0}[\frac{R^{3}}{3}-\frac{R^{2}}{2R}]=\pi J_{0}[\frac{R^{3}}{3}-\frac{R^{2}}{2}]](https://tex.z-dn.net/?f=I_%7BT%7D%3D%5Cpi%20J_%7B0%7D%20%5Cint%5Climits%5ER_0%20%7Br%5E%7B2%7D%281-r%2FR%29%7D%20%5C%2C%20dr%20%3D%20%5Cpi%20%20%20J_%7B0%7D%5B%5Cfrac%7BR%5E%7B3%7D%7D%7B3%7D-%5Cfrac%7BR%5E%7B2%7D%7D%7B2R%7D%5D%3D%5Cpi%20J_%7B0%7D%5B%5Cfrac%7BR%5E%7B3%7D%7D%7B3%7D-%5Cfrac%7BR%5E%7B2%7D%7D%7B2%7D%5D)
And it is only necessary to replace J0 and R.
I hope this is useful for you
regards
2. Wheel and axle
3.fulcrum
6.wheel and axle (I think) hope this helps!
Answer:
7.8x10-12N
Explanation:
We know that
Magnetic force = F = qVB
And
Also Kinetic energy K.E is
E = (1/2)mV²
So making v subject
V = √(2E / m)
And
E = KE = 2MeV
= 2 × 106 eV
= 2 × 106 × 1.6 × 10–19 J
= 3.2 × 10–13 J
And then
V= √2x3.2E-13/1.6E-27
1.9E7m/s
Given that
mass of proton = 1.6 × 10–27 kg,
Magnetic field strength B = 2.5 T.
So F= qBv sinစ
=
So F = 1.6 × 10–19 × 2.5 × 1.9 x10^7 x sin 90°
= 7.8 x 10^-12N