How to calculate the ampere force impulse
Publish: 2021-05-02 21:46:25
1. F = bil (magnetic inction line, current direction and Ampere force direction are perpendicular, otherwise they must be decomposed orthogonally) f: Ampere force B: magnetic inction intensity I: current intensity through wire L: effective length of cutting magnetic inction line
2. F=BIL The direction is left-handed. Can be asked
3. F = bil (bil must be vertical)
4. F = bil (magnetic inction line, current direction and Ampere force direction are perpendicular, otherwise they must be decomposed orthogonally) f: Ampere force B: magnetic inction intensity I: current intensity through wire L: effective length of cutting magnetic inction line
5. Let's adopt a pure hand look at the model that is not subject to external force. For all the models that are directly proportional to the velocity f = kV, I = ∫ FDT = ∫ kvdt = k ∫ VDT = KX
in this problem, f = BLV, k = BL, I = BLX
in this problem, f = BLV, k = BL, I = BLX
6. The formula of Ampere force impulse blq = MV cannot be used directly in college entrance examination. Because it's not a formula, it's just a method.
7. The one upstairs is wrong, impulse MVO & gt; mVt
the position of the wire frame remains unchanged, but the ampere force does negative work in the whole process, and the energy conservation can be obtained, so VO & gt; Vt From the time point of view, the ampere force and gravity are in the same direction in the rising process, but opposite in the falling process, so Ao & gt; At, and to & lt; is known from H = 1 / 2at2; TT, we can see that B is wrong. Average velocity of VO & gt; For VT, fo & gt is known from F = b2l2v / R; Ft (ampere force), and the upper and lower stages of Ampere force have done the same displacement, so a is wrong. When there is ampere force in the rising stage, take out a small segment of S1, which can be regarded as a uniform acceleration motion. The change of V is linear with the change of F (equivalent to changing V in V-T diagram multiplied by coefficient b2l2 / R into F-T diagram). The impulse of FT is equal to the area of F-T, that is, the area of v-t. when multiplying a coefficient, that is, displacement S1, the total displacement s can be obtained by integration. In the rising and falling stage, the displacement s of Ampere force is the same, so it is equivalent to impulse, C is opposite. I = BLV, so D is right
if you come across a similar question in the college entrance examination (generally not so difficult), don't spend too much time. You'd better choose less, or you may lose the big because of the small
if you can draw a picture, you'd better draw a picture. It's easy to understand, so you need to understand the meaning of a picture, such as what the area of V-T represents.
the position of the wire frame remains unchanged, but the ampere force does negative work in the whole process, and the energy conservation can be obtained, so VO & gt; Vt From the time point of view, the ampere force and gravity are in the same direction in the rising process, but opposite in the falling process, so Ao & gt; At, and to & lt; is known from H = 1 / 2at2; TT, we can see that B is wrong. Average velocity of VO & gt; For VT, fo & gt is known from F = b2l2v / R; Ft (ampere force), and the upper and lower stages of Ampere force have done the same displacement, so a is wrong. When there is ampere force in the rising stage, take out a small segment of S1, which can be regarded as a uniform acceleration motion. The change of V is linear with the change of F (equivalent to changing V in V-T diagram multiplied by coefficient b2l2 / R into F-T diagram). The impulse of FT is equal to the area of F-T, that is, the area of v-t. when multiplying a coefficient, that is, displacement S1, the total displacement s can be obtained by integration. In the rising and falling stage, the displacement s of Ampere force is the same, so it is equivalent to impulse, C is opposite. I = BLV, so D is right
if you come across a similar question in the college entrance examination (generally not so difficult), don't spend too much time. You'd better choose less, or you may lose the big because of the small
if you can draw a picture, you'd better draw a picture. It's easy to understand, so you need to understand the meaning of a picture, such as what the area of V-T represents.
8. F=ILB
=E/R*LB
=BLv/R*LB
=B²L²v/R
所以冲量为：Ft=B²L²vt/R
=E/R*LB
=BLv/R*LB
=B²L²v/R
所以冲量为：Ft=B²L²vt/R
9. Using the momentum theorem, I (sum) = MV, where V is the instantaneous velocity at 2S. If you have learned definite integral, you can use definite integral to solve this problem. If you are interested in how to solve it, you can ask. If you are not interested, you can adopt the answer from upstairs.
Hot content