Practice Lorentz force law with force due to electric & magnetic field, Learn Relation between electric and magnetic field and direction of electric force and magnetic force on positive and negative charge.
The force experienced by charge due to the presence of electric field \(\vec E\) is \(\vec F_E = q \, \vec E\).
The force experienced by charge due to the presence of magnetic field \(\vec B\) is \(\vec F_B = q \, (\vec v×\vec B) \)
The force experienced by charge due to the presence of both electric and magnetic field is \(\vec F= q \,\vec E + q \,\vec v × \vec B\)
The combined force is known as Lorentz force.
?Here, \(\vec v,\vec B \) and \(\vec E\) are mutually perpendicular to each other.
(1) When there is positive charge :
(2) When there is negative charge :
If \(\vec B\) is perpendicular to \(\vec v\), then
Magnetic Force \(\to\;\vec F_B=q\,\left(\vec v×\vec B\right)\)
\(\vec F_B=q\,vB\,sin\,90°\)
\(\vec F_B=q\,vB\) (upward direction)
Electric Force \(\to\;\vec F_E=q\,\vec E\) (Downward direction)
[\(\vec B,\;\vec v\;\&\;\vec E\) are mutually perpendicular]
\(\Sigma F_{net}=0\)
Taking upward as positive direction,
\(\vec F_B+\vec F_E=0\)
\(q\,vB-q\,E=0\)
\(B=\dfrac{E}{v}\)
A \(\dfrac{B_0}{E_0}\,m/s\)
B \({E_0}\,{B_0}\,m/s\)
C \(\dfrac{E_0}{B_0}\,m/s\)
D \((B_0+E_0)\,m/s\)
When the charge is positively charged :
The velocity of charge is along \(x\)-axis. The electric field is along \((-y)-\) axis while the magnetic field is along \((-z)-\) axis, as shown in figure.
Case 1: If the value of magnetic field and electric field is such that
\(\vec F_B>\vec F_E\)
then, initial deviation will be along \(y-\) axis.
Case 2: If the value of electric field and magnetic field is such that,
\(\vec F_E>\vec F_B\)
then, initial deviation will be along \((-y)-\) axis.
Case 3: If electric field > magnetic field,
but \(\vec F_E<\vec F_B\)
then, initial deviation will be along \(y-\) axis.
When the charge is negatively charged :
Case 1: If the value of magnetic field and electric field is such that
\(\vec F_B>\vec F_E\)
then, initial deviation will be along \((-y)-\) axis.
Case 2: If the value of electric field and magnetic field is such that
\(\vec F_E>\vec F_B\)
then, initial deviation will be along \(y-\) axis.
Case 3: If electric field > magnetic field
but \(\vec F_E<\vec F_B\)
then, initial deviation will be along \((-y)-\) axis.
When the charge is at rest i.e. \(v=0\) :
The force experienced by charge in presence of both magnetic and electric field is given by,
\(\vec F=q\,\vec E+q\,(\vec v×\vec B)\)
\(\because\;v=0\)
\(\vec F=q\,\vec E+0\)
\(\vec F=q\,\vec E\)
In this condition, the deviation of charge from its path will be along electric field only whether electric field is greater than magnetic field or magnetic field is greater then electric field.
A along \(x-\) axis
B along \(y-\) axis
C along \((-y)-\) axis
D along \((-z)-\) axis
\(T=\dfrac{2\pi\, m}{q\,B}\)
The acceleration along \(y-\) axis is given by \(a_y=\dfrac{q\,E}{m}=\dfrac{F_y}{m}\)
By third law of motion,
\(y=u_y\,T+\dfrac{1}{2}a_y\,(T)^2\)
Here, \(y\) is displacement along \(y-\) axis
\(u_y\) is speed along \(y-\) axis
\(T\) is time period
\(a_y\) is acceleration along \(y-\) axis
pitch \(=u_yT+\dfrac{1}{2}a_y\,(T)^2\)
\(\because\;u_y=0\)
\(\therefore\;P_1=\dfrac{1}{2}\,\dfrac{qE}{m}\left(\dfrac{2\pi\,m}{qB}\right)^2\)
\(P_1=\dfrac{1}{2}\,\dfrac{qE}{m}\left[\dfrac{4\pi^2\,m^2}{q^2B^2}\right]\)
\(P_1=\dfrac{2\pi^2\,mE}{q\,B^2}\)
\(P_1+P_2=\dfrac{1}{2}a_y\,(2T)^2\)
\(P_1+P_2=\dfrac{1}{2}\,\dfrac{q\,E}{m}\,\left[\dfrac{2×2\pi\,m}{q\,B}\right]^2\)
\(P_1+P_2=\dfrac{1}{2}\,\dfrac{q\,E}{m}\,\dfrac{16\pi^2\,m^2}{q^2\,B^2}\)
\(P_1+P_2=\dfrac{8\pi^2\,mE}{q\,B^2}\)
putting the value of \(P_1\),
\(\dfrac{2\pi^2\,mE}{q\,B^2}+P_2=\dfrac{8\pi^2\,mE}{q\,B^2}\)
\(P_2=\dfrac{6\pi^2\,mE}{q\,B^2}\)
\(P_1+P_2+P_3=\dfrac{1}{2}a_y\,(3T)^2\)
\(P_1+P_2+P_3=\dfrac{1}{2}\,\dfrac{q\,E}{m}×9T^2\)
\(P_1+P_2+P_3=\dfrac{1}{2}\,\dfrac{q\,E}{m}×9\,\left[\dfrac{4\pi^2m^2}{q^2B^2}\right]\)
\(P_1+P_2+P_3=\dfrac{18\pi^2m\,E}{q\,B^2}\)
putting the value of \(P_1+P_2\),
\(P_3=\dfrac{18\pi^2m\,E}{q\,B^2}-\dfrac{8\pi^2\,mE}{q\,B^2}\)
\(P_3=\dfrac{10\pi^2m\,E}{q\,B^2}\)
\(P_1:P_2:P_3=\dfrac{2\pi^2m\,E}{q\,B^2}:\dfrac{6\pi^2m\,E}{q\,B^2}:\dfrac{10\pi^2m\,E}{q\,B^2}\)
\(P_1:P_2:P_3=1:3:5\)
The pitch of the helical path increases in similar manner.
A \(4\,m\)
B \(2\,m\)
C \(5.25\,m\)
D \(3.15\,m\)
If \(\vec B\) is perpendicular to \(\vec v\), then
Magnetic Force \(\to\;\vec F_B=q\,\left(\vec v×\vec B\right)\)
\(\vec F_B=q\,vB\,sin\,90°\)
\(\vec F_B=q\,vB\) (upward direction)
Electric Force \(\to\;\vec F_E=q\,\vec E\) (Downward direction)
[\(\vec B,\;\vec v\;\&\;\vec E\) are mutually perpendicular] ?
\(\Sigma F_{net}=0\)
Taking upward as positive direction,
\(\vec F_B+\vec F_E=0\)
\(q\,vB-q\,E=0\)
\(B=\dfrac{E}{v}\)
A \(-51\,\hat k\;{V}/{m}\)
B \(12\,(-\hat k)\;{V}/{m}\)
C \(20\,(-\,\hat k)\;{V}/{m}\)
D \(-14\,\hat k\;{V}/{m}\)