Informative line

Presence Of Both E And B

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.

Lorentz Force Law

  • Consider a charge, moving with velocity \(\vec v\) in the presence of both an electric field \(\vec E\) and a magnetic field \(\vec B\).
  • Force due to electric field :

        The force experienced by charge due to the presence of electric field \(\vec E\) is  \(\vec F_E = q \, \vec E\).

  • Force due to magnetic field :

The force experienced by charge due to the presence of magnetic field \(\vec B\)  is \(\vec F_B = q \, (\vec v×\vec B) \) 

  • Lorentz Force : 

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.

  • Direction of electric force \((\vec F_E)\) and magnetic force \((\vec F_B)\) on positive and negative charge:

?Here, \(\vec v,\vec B \) and \(\vec E\) are mutually perpendicular to each other.

(1)  When there is positive charge :

  • The  direction of magnetic force is determined  by right hand thumb rule .
  • The direction of electric force is along the direction of electric field .

   (2) When there is negative charge :  

  • The  direction  of magnetic force is opposite  to the thumb, on applying right hand thumb rule.
  • The direction of electric force is opposite to the direction of electric field.

Illustration Questions

Which of the following option is correct ?

A

B

C

D

×

Option (A)  is incorrect  because the direction of magnetic force is not according to right hand thumb rule and direction of electric force is along electric field. 

image

Option (B)  is incorrect  because the direction of electric force is not along electric field. 

image

Option (C) is correct because the direction of electric force  is along electric field and direction of magnetic force is according to right hand thumb rule.

image

Option (D) is incorrect because the direction of magnetic force  is not  according to right hand thumb rule.

image

Which of the following option is correct ?

A image
B image
C image
D image

Option C is Correct

Relation Between Electric and Magnetic Field

  • For positive 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] 

  • The condition for which the particle can move both in magnetic and electric field undeviated is,

\(\Sigma F_{net}=0\)

Taking upward as positive direction,

\(\vec F_B+\vec F_E=0\)

\(q\,vB-q\,E=0\)

\(B=\dfrac{E}{v}\)

Illustration Questions

If a particle of mass \(m\) and charge \(q\) is moving in magnetic field of \(\vec B=B_0\,\hat j\;T\) and electric field \(\vec E=E_0(-\hat k)\,{V}{m^{-1}}\) with velocity \(\vec v=v_0\;\hat i\,m/s\) . Find the speed of the particle if the particle moves undeviated.

A \(\dfrac{B_0}{E_0}\,m/s\)

B \({E_0}\,{B_0}\,m/s\)

C \(\dfrac{E_0}{B_0}\,m/s\)

D \((B_0+E_0)\,m/s\)

×

Magnetic field, \(\vec B=B_0\,\hat j\; T\)

Electric field, \(\vec E=E_0(-\hat k) \;Vm^{-1}\)

Velocity of particle,  \(\vec v=v_0\,\hat i\; ms^{-1}\)

The magnetic force experienced by particle,

\(\vec F_B=q\,(\vec v×\vec B)\)

\(\vec F_B=q\,[(v_0\,\hat i)×(B_0\,\hat j)]\)

\(\vec F_B=q\,v_0\,B_0(\hat k)\)

The electric force experienced by particle,

\(\vec F_E=q\,\vec E\)

\(\vec F_E=q\,E_0(-\hat k)\)

The condition in which particle moves undeviated is

 \(\vec F_B+\vec F_E=0\)

\(q\,v_0\,B_0\,\hat k=+q\,E_0(+\hat k)\)

\(q\,v_0\,B_0=q\,E_0\)

\(v_0=\dfrac{E_0}{B_0}\;m/s\)

If a particle of mass \(m\) and charge \(q\) is moving in magnetic field of \(\vec B=B_0\,\hat j\;T\) and electric field \(\vec E=E_0(-\hat k)\,{V}{m^{-1}}\) with velocity \(\vec v=v_0\;\hat i\,m/s\) . Find the speed of the particle if the particle moves undeviated.

A

\(\dfrac{B_0}{E_0}\,m/s\)

.

B

\({E_0}\,{B_0}\,m/s\)

C

\(\dfrac{E_0}{B_0}\,m/s\)

D

\((B_0+E_0)\,m/s\)

Option C is Correct

Calculation of Electric Field for which the Charge goes Undeviated

  • For positive 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]  ?

  • The condition for which the particle can move both in magnetic and electric field undeviated is,

\(\Sigma F_{net}=0\)

Taking upward as positive direction,

\(\vec F_B+\vec F_E=0\)

\(q\,vB-q\,E=0\)

\(B=\dfrac{E}{v}\)       

Illustration Questions

If a point charge \(q=1\,\mu \,C\) is moving with velocity \(v=3\,\hat i\,m/s\) in electric field \(\vec E\) and magnetic field \(\vec B=4\,\hat j\,T\), then find the value of electric field for which point charge goes undeviated.

A \(-51\,\hat k\;{V}/{m}\)

B \(12\,(-\hat k)\;{V}/{m}\)

C \(20\,(-\,\hat k)\;{V}/{m}\)

D \(-14\,\hat k\;{V}/{m}\)

×

Charge on point charge, \(q=1\,\mu \,C\)

Velocity of point charge, \(v=3\,\hat i\,m/s\)

Magnetic field, \(\vec B=4\,\hat j\,T\)

For charge to go undeviated,

\(\Sigma F_{net}=0\)

\(\vec F_B+\vec F_E=0\)

\(q\,\vec E=-q(\vec v×\vec B)\)

\(\vec E=-(\vec v×\vec B)\)

\(\vec E=-(3\,\hat i×4\hat j)\)

\(=12(\,-\hat k\;)\,V/m\)

If a point charge \(q=1\,\mu \,C\) is moving with velocity \(v=3\,\hat i\,m/s\) in electric field \(\vec E\) and magnetic field \(\vec B=4\,\hat j\,T\), then find the value of electric field for which point charge goes undeviated.

A

\(-51\,\hat k\;{V}/{m}\)

.

B

\(12\,(-\hat k)\;{V}/{m}\)

C

\(20\,(-\,\hat k)\;{V}/{m}\)

D

\(-14\,\hat k\;{V}/{m}\)

Option B is Correct

Path of Charged Particle when Magnetic and Electric Field are Parallel

  • Consider a charged particle of mass \(m\) and charge \(q\), is moving with velocity \(v\), in presence of both magnetic and electric field.
  • The magnetic and electric field are parallel to each other.
  • The velocity of charged particle is perpendicular to the magnetic field, as shown in figure.

  • As \(\vec v\) is perpendicular to \(\vec B\), thus, the motion of particle should be circular.
  • But due to the presence of electric field, the acceleration of charged particle will be along \(\vec E\).
  • This situation can be understood easily with the help of Cartesian coordinate system, as shown in figure.
  • Assume that the direction of velocity of charged particle is along \(x\)-axis and magnetic and electric field are along \(y\)-axis.
  • Magnetic and electric field are parallel to each other.

  • The trajectory of the particle can be explained as follows :
  1. If only magnetic field would present and \(\vec v\) is perpendicular to \(\vec B\), then the motion would be circular.
  2. But due to the presence of electric field, the charged particle will attain some acceleration along the direction of electric field.
  3. Therefore, the path of charged particle will be in the form of helix, with increasing pitch, along the direction of electric field.

Illustration Questions

Which path is correct for the particle moving as shown in figure?

A

B

C

D

×

Option (A) is incorrect because the helical path of particle is along \(x\)-axis.

image

Option (B) is correct because the helical path of particle, with increasing pitch, is along \(y\)-axis.

image

Option (C) is incorrect because the helical path of particle is along \(z\)-axis.

image

Option (D) is incorrect because the helical path of particle is along \((-z)-\) axis.

image

Which path is correct for the particle moving as shown in figure?

image
A image
B image
C image
D image

Option B is Correct

Deviation in the Path of a Charge

  • Consider a charge of mass \(m\) and charge \(q\) which is moving in presence of both magnetic field and electric field.

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.

Illustration Questions

A positive charge of mass \(m\) and charge \(q\) is moving with velocity \(\vec v=2\,\hat i\;m/s\) along \(x-\) axis. It enters in electric field which is along \((-y)\) axis and magnetic field is along \(z-\) axis. If \(\vec E=3\hat j\;V/m\) and \(\vec B=5\,\hat k\,T\), then what will be the initial deviation of charge?

A along \(x-\) axis

B along \(y-\) axis

C along \((-y)-\) axis

D along \((-z)-\) axis

×

Velocity of charge, \(\vec v=2\,\hat i\;m/s\)

Magnetic field, \(\vec B=5\,\hat k\;T\)

Electric field, \(\vec E=3\,\hat j\;V/m\)

The situation can be shown by following diagram-

image

The magnetic force experienced by charge :

\(\vec F_B=q\,(\vec v×\vec B)\)

\(\vec F_B=q\,(2\,\hat i×5\,\hat k)\)

\(\vec F_B=10\,q\,(-\hat j)\,N\)

The electric force experienced by charge :

\(\vec F_E=q\,\vec E\)

\(=q\,3\,(\hat j)\)

\(=3\,q\,(\hat j)\,N\)

Since, \(\left|\vec F_B\right|>\left|\vec F_E\right|\)

Therefore, initial deviation will be along \((-y)\) axis.

A positive charge of mass \(m\) and charge \(q\) is moving with velocity \(\vec v=2\,\hat i\;m/s\) along \(x-\) axis. It enters in electric field which is along \((-y)\) axis and magnetic field is along \(z-\) axis. If \(\vec E=3\hat j\;V/m\) and \(\vec B=5\,\hat k\,T\), then what will be the initial deviation of charge?

A

along \(x-\) axis

.

B

along \(y-\) axis

C

along \((-y)-\) axis

D

along \((-z)-\) axis

Option C is Correct

Pitch of the Path Covered by Charged Particle

  • Consider a charged particle of mass \(m\) and charge \(q\) is moving with velocity \(v\), in presence of both magnetic and electric field.
  • The magnetic and electric field are parallel to each other.
  • The velocity of charged particle is perpendicular to the magnetic field, as shown in figure.

  • As \(\vec v\) is perpendicular to \(\vec B\), thus, the motion of particle should be circular.
  • But due to the presence of electric field, the acceleration of charged particle will be along \(\vec E\).
  • This situation can be understood easily with the help of Cartesian coordinate system, as shown in figure.
  • Assume that the direction of velocity of charged particle is along \(x\)-axis and magnetic and electric field are along \(y\)-axis.
  • Magnetic and electric field are parallel to each other.
  • The trajectory of the particle can be explained as follows :
  1. If only magnetic field would present and \(\vec v\) is perpendicular to \(\vec B\), then the motion would be circular.
  2. But due to the presence of electric field, the charged particle will attain some acceleration along the direction of electric field.
  3. Therefore, the path of charged particle will be in the form of helix, with increasing pitch, along the direction of electric field.

  • Consider that the value of electric field is \(\vec E=E_0\,\hat j\), the value of magnetic field is \(\vec B=B_0\,\hat j\) and the value of velocity is \(v=v_0\,\hat i\).
  • Here, pitch is the displacement along \(y-\) axis in time period of one revolution.
  • The time period of one revolution is given by

\(T=\dfrac{2\pi\, m}{q\,B}\)

The acceleration along \(y-\) axis is given by \(a_y=\dfrac{q\,E}{m}=\dfrac{F_y}{m}\)

  • The value of first pitch :

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 \(=u_yT+\dfrac{1}{2}a_y\,(T)^2\)

\(\because\;u_y=0\)

\(\therefore\;P_1=\dfrac{1}{2}\,\dfrac{qE}{m}\left(\dfrac{2\pi\,m}{qB}\right)^2\)

\(P_1=\dfrac{1}{2}\,\dfrac{qE}{m}\left[\dfrac{4\pi^2\,m^2}{q^2B^2}\right]\)

\(P_1=\dfrac{2\pi^2\,mE}{q\,B^2}\)

  • Up to second revolution, time period is \(2\,T\).

  • The value of second pitch :

\(P_1+P_2=\dfrac{1}{2}a_y\,(2T)^2\)

\(P_1+P_2=\dfrac{1}{2}\,\dfrac{q\,E}{m}\,\left[\dfrac{2×2\pi\,m}{q\,B}\right]^2\)

\(P_1+P_2=\dfrac{1}{2}\,\dfrac{q\,E}{m}\,\dfrac{16\pi^2\,m^2}{q^2\,B^2}\)

\(P_1+P_2=\dfrac{8\pi^2\,mE}{q\,B^2}\)

putting the value of \(P_1\),

\(\dfrac{2\pi^2\,mE}{q\,B^2}+P_2=\dfrac{8\pi^2\,mE}{q\,B^2}\)

\(P_2=\dfrac{6\pi^2\,mE}{q\,B^2}\)

  • Up to third revolution, time period is \(3\,T\).

  • The value of third pitch :

\(P_1+P_2+P_3=\dfrac{1}{2}a_y\,(3T)^2\)

\(P_1+P_2+P_3=\dfrac{1}{2}\,\dfrac{q\,E}{m}×9T^2\)

\(P_1+P_2+P_3=\dfrac{1}{2}\,\dfrac{q\,E}{m}×9\,\left[\dfrac{4\pi^2m^2}{q^2B^2}\right]\)

\(P_1+P_2+P_3=\dfrac{18\pi^2m\,E}{q\,B^2}\)

putting the value of \(P_1+P_2\),

\(P_3=\dfrac{18\pi^2m\,E}{q\,B^2}-\dfrac{8\pi^2\,mE}{q\,B^2}\)

\(P_3=\dfrac{10\pi^2m\,E}{q\,B^2}\)

  • Ratio of first, second and third pitch :

\(P_1:P_2:P_3=\dfrac{2\pi^2m\,E}{q\,B^2}:\dfrac{6\pi^2m\,E}{q\,B^2}:\dfrac{10\pi^2m\,E}{q\,B^2}\)

\(P_1:P_2:P_3=1:3:5\)

The pitch of the helical path increases in similar manner.

Illustration Questions

A charged particle of mass \(m=1\,\,\mu g\) and charge \(q=1\,\,\mu C\) is moving with velocity \(\vec v=4\,\hat i\;m/s.\) It enters in magnetic and electric field of \(\vec B=5\,\hat jT\) and \(\vec E =4\,\hat j\;V/m\), respectively. Find the pitch of the path in first revolution.

A \(4\,m\)

B \(2\,m\)

C \(5.25\,m\)

D \(3.15\,m\)

×

Mass of charged particle, \(m=1\,\mu g\)

Charge on charged particle, \(q=1\,\mu C\)

Velocity of charged particle, \(\vec v=4\,\hat i\;m/s\)

Magnetic field, \(\vec B=5\,\hat j \,T\)

Electric field, \(\vec E=4\,\hat j\;V/m\)

The pitch of the path in first revolution :

\(P_1=\dfrac{1}{2}\,a_y\,T^2\)

\(P_1=\dfrac{1}{2}\left(\dfrac{qE}{m}\right)\,\left(\dfrac{4\pi^2m^2}{q^2B^2}\right)\)

\(P_1=\dfrac{2\pi^2mE}{q\,B^2}\)

\(P_1=\dfrac{2(3.14)^21×10^{-6}×4}{1×10^{-6}×25}\)

\(P_1=3.15\,m\)

A charged particle of mass \(m=1\,\,\mu g\) and charge \(q=1\,\,\mu C\) is moving with velocity \(\vec v=4\,\hat i\;m/s.\) It enters in magnetic and electric field of \(\vec B=5\,\hat jT\) and \(\vec E =4\,\hat j\;V/m\), respectively. Find the pitch of the path in first revolution.

A

\(4\,m\)

.

B

\(2\,m\)

C

\(5.25\,m\)

D

\(3.15\,m\)

Option D is Correct

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