Informative line

Presence Of Only Magnetic Field

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.

Magnetic Force on a Moving Charge

  • When a charge moves in presence of magnetic field, then the magnetic field exerts a force on it, which is known as magnetic force.
  • The magnetic force depends on both the vectors, i.e., magnetic field and velocity of moving charge.

  • Consider a positive charge which is moving along x-axis with velocity v.
  • The magnetic field is along y-axis.

Direction of Magnetic Force

  • The force on the moving charge will be along positive z-axis.

Right Hand Rule

  • Right hand rule is used to determine the direction of magnetic force.
  • It states that, curl the fingers of your right hand from \(\vec v\)  towards \(\vec B\) with your thumb extending outwards and perpendicular to the plane of \(\vec v\) and \(\vec B\) as shown in figure.
  • The direction of magnetic field is shown by your thumb.

  • When \(\vec v\) makes an angle \(\theta\) with \(\vec B\) :

          \(\vec F = q\,(\vec v × \vec B)\)

  • Using right hand rule, curl the fingers of your right hand from \(\vec v\) towards \(\vec B\) with your thumb extending outwards and perpendicular to the plane of \(\vec v\) and \(\vec B\).
  • The direction in which thumb is pointing, is the direction of magnetic force.

Illustration Questions

Choose the correct direction of force.

A

B

C

D

×

Option (A) is incorrect because no force will act on charge as charge is at rest.

Option (B) is correct because by using right hand thumb rule, the direction of force will be out of paper.

Option (C) is incorrect because by using right hand thumb rule , the direction of force will be opposite to as shown in option.

Option (D) is incorrect because force acting on charge will be zero.

\(\vec F=q\, (\vec v × \vec B)\)

\(\vec F=q\, vB\; sin\theta\)

\(\because\, \theta\) is 0°

\(\vec F=q\, vB\; sin\;0°\)

\(\vec F=0\)

Choose the correct direction of force.

A image
B image
C image
D image

Option B is Correct

Motion of a Charged Particle in Uniform Magnetic Field

  • Consider a charge \(q\),  injected into uniform magnetic field with velocity \(v\), as shown in figure.

Direction of Magnetic Force on the Charge \(q\)

  • According to the right hand thumb rule, the direction of magnetic force on charge \(q\) is perpendicularly upward to the velocity \(\vec v\).

Circular Motion of Charge q

  • The magnetic force acting on the charge \(q\) causes uniform circular motion of charge q , as shown in figure.

Effect of magnetic force on velocity

  • The magnetic force changes the direction of velocity, but not speed.

Generally,  

  1. When the velocity of charge makes some angle \(\theta\), then it means that it is neither parallel nor perpendicular to the field.
  2. The component of velocity \(\vec v\), which is parallel to \(\vec B\), is not affected by the field and is responsible for the forward motion of charge along \(\vec B\).
  3. Component of \(\vec v\) which is perpendicular to \(\vec B\), is responsible for circular motion.
  4. Thus, the charged particle spirals around the magnetic field lines forming a helical trajectory , as shown in figure.

Illustration Questions

Identify the motion of a charge, when it enters in uniform magnetic field making some angle \(\theta\) with magnetic field lines.

A

B

C

D

×

Option (A) is correct because the parallel component of velocity moves the charge in forward direction i.e. along \(\vec B\) while the perpendicular component of velocity moves the charge in circular motion. Thus, the path of the charge becomes helical.

Option (B) is incorrect because the charge moves in helical path not in circular path.

Option (C) is incorrect because the charge moves in helical path not in triangular path.

Option (D) is incorrect because the charge moves in helical path not in zig zag path.

Identify the motion of a charge, when it enters in uniform magnetic field making some angle \(\theta\) with magnetic field lines.

A image
B image
C image
D image

Option A is Correct

Calculation of Magnetic Force on a Charged Particle moving in a Uniform Magnetic Field

  • Consider a charged particle moving with velocity \(\vec v\) in uniform magnetic field.
  • The moving charged particle will experience some magnetic force \(\vec F_B\).
  • The magnitude of \(\vec F_B\) is directly proportional to 
  1. The charge \(q\) on charged particle.
  2. The speed \(\vec v\) of charged particle.
  3. The magnitude of magnetic field \(\vec B\).
  4. The sine of angle between the velocity  \(\vec v\) and magnetic field \(\vec B\).

Mathematically,

           \(|\vec F_B|=|q|\,\vec v\,B\,sin\theta\)

           \(\vec F_B=q (\vec v × \vec B)\)

  • Consider a positive charge which is moving along x-axis with velocity \(\vec v\).
  • The magnetic field is along y-axis.

Direction of Magnetic force

  • The force on the moving charge will be along positive z-axis.

Right Hand Rule

  • Right hand rule is used to determine the direction of magnetic force.
  • It states that, curl the fingers of your right hand from \(\vec v\) towards \(\vec B\) with your thumb extending outwards and perpendicular to the plane of \(\vec v\) and \(\vec B\) as shown in figure.
  • The direction of magnetic field is shown by your thumb.

  • When \(\vec v\) makes an angle \(\theta\) with \(\vec B\) 

           \(\vec F = q\,(\vec v × \vec B)\)

  • Using right hand rule, curl the fingers of your right hand from  \(\vec v\) towards \(\vec B\) with your thumb extending outwards and perpendicular to the plane of \(\vec v\) and \(\vec B\).
  • The direction in which thumb is pointing, is the direction of magnetic force.

Illustration Questions

An electron moves in a magnetic field of \(|\vec B|=\)0.030 T with velocity \(|\vec v|=\) 5 × 106 m/s making an angle \(\theta =\) 37° with the field. Calculate the magnetic force acting on moving electron. \(\Big(sin37°=\dfrac {3}{5}\Big)\)

A 1.44 × 10–14 N

B 3.8 × 10–16 N

C 2.8 × 10–14 N

D 4 × 10–14 N

×

Charge on electron, \(q\) = 1.6 × 10–19 C

Velocity of electron, \(|\vec v|=\) 5 × 106 m/s 

Magnitude of magnetic field, \(|\vec B|=\)0.030 T

Angle between \(\vec v\) and \(\vec B\),  \(\theta =\) 37° 

Magnetic force acting on moving electron is 

\(F_B=|q|\,v\,B\,sin\theta\)

      = 1.6 × 10–19 × 5× 106 × 0.030 × sin 37°

      = 1.44 × 10–14 N

An electron moves in a magnetic field of \(|\vec B|=\)0.030 T with velocity \(|\vec v|=\) 5 × 106 m/s making an angle \(\theta =\) 37° with the field. Calculate the magnetic force acting on moving electron. \(\Big(sin37°=\dfrac {3}{5}\Big)\)

A

1.44 × 10–14 N

.

B

3.8 × 10–16 N

C

2.8 × 10–14 N

D

4 × 10–14 N

Option A is Correct

Calculation of Radius and Time period of Charged Particle in Uniform Circular Motion

  • Consider, a charged particle is projected perpendicular to the magnetic field with velocity \(\vec v\) as shown in figure.

  • The velocity of charged particle is perpendicular to the magnetic field.

Path of charged particle when \(\vec F_B\) is perpendicular to \(\vec v\)

  • When magnetic force is perpendicular to velocity, then the path of particle is a circle.

Magnetic force acting on charged particle when \(\vec v\) is perpendicular to \(\vec B\)

  • Since, \(\vec v\) is perpendicular to \(\vec B\)  therefore, \(\theta\) = 90°.

         Magnetic force will be 

         \(F_B=q\,v\,B sin90°\)

         \(F_B=q\,v\,B\) ...(1)

Radius of the circular motion

         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 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.

  • It is clear from equation (4) that the time period of circular motion is independent of speed of the charged particle and radius of the circular motion.

It's important:

  • If the moving charged particle makes an angle \(\theta\) with the magnetic field, then its path will be in the form of helix.

  • In this case, resolve the velocity into two components i.e., \(v_{\parallel}\) and \(v_{\perp}\)

          where,

          \(v_{\parallel}\) is the velocity component along \(\vec B\) 

          \(v_{\perp}\)is the velocity component perpendicular to \(\vec B\).

  • The component \(v_{\parallel}\) will remain unchanged while the component \(v_{\perp}\)will change.
  • The radius of circular motion will be given by

          \(r=\dfrac {m\,v_{\perp}}{q\,B}\)

Illustration Questions

If the charge of the particle moving with constant speed in uniform circular motion is doubled, then the radius of the circular motion will 

A become half

B remain constant

C become double

D become triple

×

Since, \(r\propto\dfrac {1}{q}\)

Therefore, when charge is doubled, then the radius will become half

Hence, option (A) is correct

If the charge of the particle moving with constant speed in uniform circular motion is doubled, then the radius of the circular motion will 

A

become half

.

B

remain constant

C

become double

D

become triple

Option A is Correct

Calculation of Time Spent by Charged Particle in Uniform Magnetic Field when it enters with some Angle

  • Consider a charged particle which enters in a uniform magnetic field with some angle  \(\theta\), as shown in figure.

  • Here, the direction of magnetic field is perpendicular i.e., inside the plane of paper.
  • The charged particle enters in the magnetic field through AF with velocity \(v\).
  • By geometry,

          \(\angle FCD=\theta+\theta = \theta' \)

         \(arc\ FD = radius \times \theta'\)

          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 by Charged Particle in Uniform Magnetic Field

 \(Time \,spent=\dfrac {\text {arc } FD}{\text {velocity}}\)

   \(T=\dfrac {\dfrac {m\,v}{q\,B}×\theta'}{v}\)

   \(T=\dfrac{m\,\theta'}{q\,B}\)

   \(\because \theta'=2\theta\)

   \(T=\dfrac{2m\,\theta}{q\,B}\) 

where  \(\theta\) must be in radian.

Illustration Questions

If a particle of mass \(m\) and charge \(q\) enters in a uniform magnetic field \(\vec B\) with velocity \(\vec v\) at an angle  \(\theta=\dfrac{\pi}{4}\) , as shown in figure, then find the time spent by the particle in the uniform magnetic field.

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

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

C \(\dfrac {3\,\pi\,m}{q\,B}\)

D \(\dfrac {4\,\pi\,m}{q\,B}\)

×

\(m\)=Mass of particle

\(q\)=Charge on particle

\(\vec v\)=Velocity of particle

Angle \(\theta =\)45°

Time spend by particle is given by

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

\(T=\dfrac {2m}{q\,B}\left ( \dfrac {\pi}{4}\right)\)

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

If a particle of mass \(m\) and charge \(q\) enters in a uniform magnetic field \(\vec B\) with velocity \(\vec v\) at an angle  \(\theta=\dfrac{\pi}{4}\) , as shown in figure, then find the time spent by the particle in the uniform magnetic field.

image
A

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

.

B

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

C

\(\dfrac {3\,\pi\,m}{q\,B}\)

D

\(\dfrac {4\,\pi\,m}{q\,B}\)

Option B is Correct

Angle of Deviation of Velocity of Charged Particle

  • Consider a charged particle of mass \(m\) and charge \(q\).

Case-I

  • The charged particle enters in a uniform magnetic field \(\vec B\) with some velocity \(\vec v\) at an angle 90°,as shown in figure.

Path of charged particle

  • When charged particle enters in uniform magnetic field at right angle, then particle travels in semicircular path.
  • Also, it comes out of magnetic field at right angle.

Case-II

Angle of deviation

  • Consider a particle of charge q and mass m enters in a region of uniform magnetic field \(\vec B\).

  • The particle enters in magnetic field at right angle with velocity \(v\).
  • The width of region of magnetic field is d.
  • It deviates from its original path and traces a curved path along ADE as shown in figure.
  • While tracing the curved path it leaves the region of magnetic field at an angle \(\delta\) with horizontal.
  • The radius of curvature of curved path is R and its center is C, as shown in figure. 
  • From \(\Delta OCD\),

\(sin\,\delta=\dfrac {OD}{CD}\)

\(sin\,\delta=\dfrac {d}{R}\)

\(\delta=sin^{-1} \left (\dfrac {d}{R}\right)\)

By above formula, angle of deviation of a particle leaving the magnetic field can be determined.

Illustration Questions

A charged particle of mass \(m\) and charge \(q\) is projected at right angle into a region of uniform magnetic field \(\vec B\). What will be the angle of deviation when it comes out of the region of uniform magnetic field of width d = 3 m with radius R = 5 m. \([ sin\,37° =\dfrac{3}{5}\,,sin\,53=\dfrac{4}{5} ]\)

A 60°

B 37°

C 53°

D 30°

×

Given:

Width of magnetic field, d = 3 m

Radius of curvature of path R = 5 m

Angle of deviation

\(\delta=sin^{-1} \left (\dfrac {d}{R}\right)\)

\(=sin^{-1} \left (\dfrac {3}{5}\right)\)

\(\delta=37°\)

A charged particle of mass \(m\) and charge \(q\) is projected at right angle into a region of uniform magnetic field \(\vec B\). What will be the angle of deviation when it comes out of the region of uniform magnetic field of width d = 3 m with radius R = 5 m. \([ sin\,37° =\dfrac{3}{5}\,,sin\,53=\dfrac{4}{5} ]\)

A

60°

.

B

37°

C

53°

D

30°

Option B is Correct

Motion of a Charged Particle in Uniform Magnetic Field

  • Consider a charge \(q\), is injected into uniform magnetic field with velocity \(v\), as shown in figure.

Direction of Magnetic Force on the Charge q

  • According to the right hand thumb rule, the direction of magnetic force on charge \(q\) is perpendicularly upward to the velocity \(\vec v\).

Circular Motion of Charge q

  • The magnetic force acting on the charge \(q\) causes uniform circular motion of charge \(q\), as shown in figure.

Effect of Magnetic Force on Velocity

  • The magnetic force changes the direction of velocity, but not speed.

Generally

  1. When the velocity of charge makes some angle \(\theta\) then it means that it is neither parallel nor perpendicular to the field.
  2. The component of velocity \(\vec v\), which is parallel to \(\vec B\), is not affected by the field and is responsible for the forward motion of charge along \(\vec B\).
  3. The component of \(\vec v\) which is perpendicular to \(\vec B\), is responsible for circular motion.
  4. Thus, the charged particle spirals around the magnetic field lines forming a helical trajectory, as shown in figure.

Radius of Helical Path

Illustration Questions

If a charged particle enters in a magnetic field at an angle of 53°, then choose the path and pitch of journey of particle.

A

B

C

D

×

Option (A) is correct because the path of charged particle enters in magnetic field at an angle of 53° is helical and pitch is the distance covered by the charged particle along the magnetic field during time period.

Option (B) is incorrect because the pitch of helical path is not the length of diameter of circular motion.

Option (C) is incorrect because the helical path of charged particle is not along the magnetic field.

Option (D) is incorrect because the path of charged particle is not circular in magnetic field. Also, the pitch is not the radius of the circle.

If a charged particle enters in a magnetic field at an angle of 53°, then choose the path and pitch of journey of particle.

A image
B image
C image
D image

Option A is Correct

Illustration Questions

A proton enters in a uniform magnetic field of B = 0.3 T with velocity v = 15 × 105 m/s. The velocity makes an angle  \(\theta =\)37° with the magnetic field. Find the radius of helical path, time period and pitch of the helix.\(\left [ cos37°=\dfrac {4}{5},\, sin37°=\dfrac {3}{5},\, m_p=1.6×10^{-27}\,kg \right ]\) 

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

×

Mass of proton, m = 1.6 × 10–27 kg

Velocity of proton, v = 15 × 105 m/s

Magnitude of magnetic field, B = 0.3 T

Angle \(\theta=37°\)

Charge on proton, q = 1.6 × 10–19 C

The radius of helical path is

\(r=\dfrac {mv_{\perp}}{q\,B}\)

\(r=\dfrac {mv\,sin\theta}{q\,B}\)

\(r=\dfrac {1.6×10^{-27}×15×10^5×sin37°}{1.6×10^{-19}×0.3}\)

\(r=\dfrac {1.6×10^{-27}×15×10^5×(3/5)}{1.6×10^{-19}×0.3}\)

\(r= {3×10^{-2}}\ m \)

 

Time period of helical path

\(T=\dfrac {2\,\pi\,r}{v_{\perp}}\)

\(T=\dfrac {2\,\pi\,r}{v\,sin\theta}\)

\(T=\dfrac {2×3.14×3×10^{-2}}{15×10^5×sin37°}\)

\(T=\dfrac {2×3.14×3×10^{-2}}{15×10^5×(3/5)}\)

\(T= {2.09×10^{-7}} sec\)

Pitch of the helix

 \(P=v_{\parallel}×T\)

P = v cos \(\theta\) × 2.09 × 10–7

P = 15 × 105 × cos 37° × 2.09 × 10–7

P = 15 × 105× (4/5) × 2.09 × 10–7

P = 2.5 × 10–1

A proton enters in a uniform magnetic field of B = 0.3 T with velocity v = 15 × 105 m/s. The velocity makes an angle  \(\theta =\)37° with the magnetic field. Find the radius of helical path, time period and pitch of the helix.\(\left [ cos37°=\dfrac {4}{5},\, sin37°=\dfrac {3}{5},\, m_p=1.6×10^{-27}\,kg \right ]\) 

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

Option C is Correct

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