Moving Charges and Magnetism Test - 40

Since B is in motion with respect to the charge so he can not see any magnetic field but A will see that. 

$$F=B  i  l  sin \theta$$
Since current carrying wire is perpendicular to magnetic field of induction 
$$\theta = 90^{\circ}$$
$$F = Bil  sin  90^{\circ} = Bil$$

Current carrying conductor will generate magnetic field in its surrounding which will have its impact on an another current carrying conductor.
Force per unit length between two current carrying conductor is given by,
$$\bigg [F = \dfrac{(\mu _{0}\times I_{1}\times I_{2})}{2\times \pi \times d}\bigg ]$$
where , $$d$$ is the distance between the conductors.
We can see from the equation that force is inversely proportional to the distance between the two conductors.

If a positively charged particle moving in east direction enters a region of uniform magnetic field directed vertically upwards, it will move in a circular path with a uniform speed.
If a charged particle is moving in a direction perpendicular to a uniform magnetic field, then its trajectory will be a circle because the force $$F=qvB$$ is always perpendicular to the velocity, and therefore centripetal.

Force per unit length between two current carrying conductor is given by ,
$$[F = \dfrac{(\mu _{0}I_{1}I_{2})}{2\pi d}]$$
so, if we doubles the current values the force will become $$4$$ times.

Force on a current carrying wire placed in a magnetic field is given by $$F = I(l \times B)$$
So when the wires are perpendicular the magnetic field produced by one of them will be along the length of the other and we know cross product of two vectors along the same direction is zero.

$$\overrightarrow{F}_{mag}= q(\overrightarrow{v} \times \overrightarrow{B})$$
Work done on the particle $$W= \int P dt$$
But, $$P = \overrightarrow{F}_{mag} .\overrightarrow{v}= q(\vec{v} \times \vec{B}).\overrightarrow{v}=0$$ since $$\overrightarrow{v} \times \overrightarrow{B} \perp \overrightarrow{v}$$
$$\therefore W=0$$.
Also, one can look at it from displacement perspective. Since the particle comes back to the same position after one cycle, $$\overrightarrow{ds}$$ is zero.
$$\Rightarrow W=0$$

The force per unit length  between two long current carrying carrying conductors is $$F=\dfrac{\mu_0 i_1i_2}{2\pi d}$$
where d is the distance between the conductors and $$i_1$$ and $$i_2$$ are the currents carried by the conductors.
The conductors attract each other if $$i_1$$ and $$i_2$$ are in the same direction. Magnetic field created by conductor 1 at the location of the conductor 2  and conductor 2 gets attracted towards conductor 1.
Since $$i_1=i_2=i$$
$$ \therefore F = \dfrac{\mu_0 i^2}{2\pi d}$$

Force on a charged particle q having velocity $$v$$ in the magnetic field(B) is = $$F_B$$ = q$$v$$B
Centripetal force on the particle moving in a circular path = $$F_C$$ = $$\frac {m{v}^2}{R}$$
Now for particle to remain in circular path = $$F_B = F_C$$
q$$v$$B = $$\frac {m{v}^2}{R}$$
q$$v_1$$B = $$\frac {m_1{v_1}^2}{R_1}$$   (For particle 1)  $$\Rightarrow$$  qB = $$\frac {m_1{v_1}}{R_1}$$                .......eq.(1)            
q$$v_2$$B = $$\frac {m_2{v_2}^2}{R_2}$$   (For particle 2)  $$\Rightarrow$$  qB = $$\frac {m_2{v_2}}{R_2}$$                .......eq.(2)
By equating eq.(1) & (2)  $$\Rightarrow$$  $$\frac {m_1{v_1}}{R_1}$$ = $$\frac {m_2{v_2}}{R_2}$$
$$(\frac {m_1}{m_2}) = (\frac {v_2}{v_1})\times(\frac {R_1}{R_2})$$                                                       ........eq.(3)
By accelerating the charged particle q with voltage (V), work done by the particle(W) = qV
W = $$\Delta$$K    (work energy theorem)
qV = $$\frac{1}{2}m{v}^2$$ - 0
$$\Rightarrow \frac {1}{2}{m_1}{v_1}^2 = qV = \frac {1}{2}{m_2}{v_2}^2$$
$$\Rightarrow {m_1}{v_1}^2 = {m_2}{v_2}^2$$
$$(\frac {v_2}{v_1}) = (\frac {m_1}{m_2})^{\frac {1}{2}}$$                                                                  ..........eq.(4)
Now from eq.(3) & (4) $$\Rightarrow$$  $$(\frac {m_1}{m_2}) = (\frac {m_1}{m_2})^{\frac {1}{2}}(\frac {R_1}{R_2})$$
$$(\frac {m_1}{m_2})^{\frac {1}{2}}$$ = $$(\frac {R_1}{R_2})$$
$$(\frac {m_1}{m_2})$$ = $$(\frac {R_1}{R_2})^{2}$$