Learn correct direction of magnetic force using right hand rule, practice to find the magnetic force on a charged particle moving in a uniform magnetic field and determining the velocity of a charged particle.
\(\vec F = q\,(\vec v × \vec B)\)
Generally,
Mathematically,
\(|\vec F_B|=|q|\,\vec v\,B\,sin\theta\)
\(\vec F_B=q (\vec v × \vec B)\)
\(\vec F = q\,(\vec v × \vec B)\)
A 1.44 × 10–14 N
B 3.8 × 10–16 N
C 2.8 × 10–14 N
D 4 × 10–14 N
Magnetic force will be
\(F_B=q\,v\,B sin90°\)
\(F_B=q\,v\,B\) ...(1)
For uniform circular motion, \(F=\dfrac {m\,v^2}{r}\) ...(2)
Comparing equation (1) and (2)
\(q\,v\,B=\dfrac {m\,v^2}{r}\)
\(r=\dfrac {m\,v}{q\,B}\) ...(3)
where, \(r\) is the radius of circular motion.
Time period is the time taken by the particle to complete one revolution.
It is given by
Time period = \(\dfrac {\text {circumference}}{\text{speed}}\)
\(T=\dfrac {2\pi\,r}{v}\)
from equation (3) \(r=\dfrac {m\,v}{q\,B}\)
\(T=\dfrac {2\,\pi \,m\,v}{v\,q\,B}\)
\(T=\dfrac {2\,\pi \,m}{q\,B}\) ....(4)
where, T is the time period of circular motion.
If the moving charged particle makes an angle \(\theta\) with the magnetic field, then its path will be in the form of helix.
where,
\(v_{\parallel}\) is the velocity component along \(\vec B\)
\(v_{\perp}\)is the velocity component perpendicular to \(\vec B\).
\(r=\dfrac {m\,v_{\perp}}{q\,B}\)
A become half
B remain constant
C become double
D become triple
\(\angle FCD=\theta+\theta = \theta' \)
\(arc\ FD = radius \times \theta'\)
Here, \(\theta'\) is angle subtended by \(arc\ FD\) at center C of circle.
In circular uniform motion,
\(r=\dfrac {m\,v}{q\,B}\)
Thus, \(arc\,\, FD=\dfrac {m\,v}{q\,B}×\theta'\)...(1)
\(Time \,spent=\dfrac {\text {arc } FD}{\text {velocity}}\)
\(T=\dfrac {\dfrac {m\,v}{q\,B}×\theta'}{v}\)
\(T=\dfrac{m\,\theta'}{q\,B}\)
\(\because \theta'=2\theta\)
\(T=\dfrac{2m\,\theta}{q\,B}\)
where \(\theta\) must be in radian.
A \(\dfrac {2\,\pi\,m}{q\,B}\)
B \(\dfrac {\pi\,m}{2\,q\,B}\)
C \(\dfrac {3\,\pi\,m}{q\,B}\)
D \(\dfrac {4\,\pi\,m}{q\,B}\)
\(sin\,\delta=\dfrac {OD}{CD}\)
\(sin\,\delta=\dfrac {d}{R}\)
\(\delta=sin^{-1} \left (\dfrac {d}{R}\right)\)
By above formula, angle of deviation of a particle leaving the magnetic field can be determined.
Generally
A r = 2 × 10–20 m, T = 5.6 × 10–24 sec, p = 2.61 × 10–15 m
B r = 3 × 10–13 m, T = 0.5 × 10–16 sec, p = 5.6 × 10–15 m
C r = 3 × 10–2 m, T = 2.09 × 10–7 sec, p = 2.5 × 10–1 m
D r = 2 × 10–13 m, T = 3 × 10–16 sec, p = 2 × 10–20 m