Question

एक समान चुंबकीय क्षेत्र में लंबवत प्रवेश करने वाले किसी आवेशित कण द्वारा प्राप्त वृत्तीय पथ की त्रिज्या का सूत्र ज्ञात कीजिए

Answer 1

The required formula is:

$R=\frac{mv\sin \theta }{qB}$

Explanation:

• Helical path is that the path of the motion of a charged particle once enters at associate angle of θ in an exceedingly magnetic flux.
• During this short tutorial, we have a tendency to make a case for the factors that cause this kind of motion.
• On a moving charged particle in an exceedingly magnetic flux, a attractive force of magnitude $F_B} } =qv B sin$θ   is acted wherever θ  is that the angle of rate $v$ vector  with the flux $B$ vector.
• This is often the most issue that makes a spiral or helical path.  
• A charged particle (say, electron) will enter a section stuffed with uniform B  either with right angle θ = 90°  or at angle θ.
• Within the former case, its path ends up in a circular path, and within the latter case, a helical path is created. Currently  we wish to answer this question: why do charged particles move in an exceedingly helical path.
• As presently as a charged particle enter a magnetic flux B with some angle θ , one will decompose its velocity into parallel and vertical parts with regard to $B$ vector  which are $v$║=$vcos$θ and the uniform magnetic flux B  will not apply any force on the charged particle (say, electron) within the parallel direction that’s F║=$qv$║$Bsin$θ.
• Thus, the charged particle continues to maneuver on the sector direction with a consistent motion (a motion within which speed and rate is constant).
• On the opposite hand, the vertical element undergoes a attractive force of magnitude F⊥=$qv$⊥$Bsin$90°
• In the attached image, those 2 on top of motions, uniform motion parallel to the sector   and uniform circular motion perpendicular to the sector B, creates the particular path of a charged particle in an exceedingly magnetic flux B that is analogous to a spring and is named a spiral or helical path.    
• Every helical path has 3 distinct characteristics as radius, period of time and pitch.
• RADIUS: The conventional force that creates a circular motion provides a force on the charged particle with a radial acceleration $a_{r} =\frac{mv^{2} }{R}$.
• Applying Newton's second law of motion and equalization the force with the attractive force we have a tendency to get a formula for radius of helical path as    

                                      $F=ma_{r}$

                                     $qvB=mv^{2} /R$

                                    $R=mv/qB$

                                    $R=mv sin \theta/qB$