A proton traveling at 42 . 0 ° with respect to the direction of the magnetic field of strength 2 . 60 mT experiences a magnetic force of 1 . 17 × 10 - 17 N . Calculate (a) the proton's speed, (b) its kinetic energy in electron-volts, and (c) its momentum. Take, mass of proton = 1 . 67 × 10 - 27 kg and sin 42 ° = 0 . 669 .

The magnetic force on a charged particle is given by F → B = q v → × B → , where v → is the velocity of the charged particle and B → is the magnetic field. The magnitude of the magnetic force on the proton (of charge + e ) is F B = e v B sin ϕ , where ϕ is the angle between v → and B → ( a ) The speed of the proton is, v = F B e B sin ϕ = 1 . 17 × 10 - 17 N 1 . 60 × 10 - 19 C 2 . 60 × 10 - 3 T sin 42 . 0 ° = 4 . 20 × 10 4 m s - 1 ( b ) The kinetic energy of the proton is, K = 1 2 m v 2 = 1 2 1 . 67 × 10 - 27 kg 4 . 20 × 10 4 m s - 1 2 = 1 . 47 × 10 - 18 J Which is equivalent to, K = 1 . 47 × 10 - 18 J 1 . 60 × 10 - 19 J eV - 1 = 9 . 18 eV From the definition of B → given by the expression F → B = q v → × B → , we see that the magnetic force F → B is always perpendicular to v → and B → . (c) The momentum of the proton is p = m v = ( 1 . 67 × 10 - 27 kg ) ( 4 . 00 × 10 4 m s - 1 ) = 7 . 01 × 10 - 23 kg m s - 1

A proton traveling at 42 . 0 ° with respect to the direction of the magnetic field of strength 2 . 60 mT experiences a magnetic force of 1 . 17 × 10 - 17 N . Calculate (a) the proton's speed, (b) its kinetic energy in electron-volts, and (c) its momentum. Take, mass of proton = 1 . 67 × 10 - 27 kg and sin 42 ° = 0 . 669 .

The magnetic force on a charged particle is given by F → B = q v → × B → , where v → is the velocity of the charged particle and B → is the magnetic field. The magnitude of the magnetic force on the proton (of charge + e ) is F B = e v B sin ϕ , where ϕ is the angle between v → and B → ( a ) The speed of the proton is, v = F B e B sin ϕ = 1 . 17 × 10 - 17 N 1 . 60 × 10 - 19 C 2 . 60 × 10 - 3 T sin 42 . 0 ° = 4 . 20 × 10 4 m s - 1 ( b ) The kinetic energy of the proton is, K = 1 2 m v 2 = 1 2 1 . 67 × 10 - 27 kg 4 . 20 × 10 4 m s - 1 2 = 1 . 47 × 10 - 18 J Which is equivalent to, K = 1 . 47 × 10 - 18 J 1 . 60 × 10 - 19 J eV - 1 = 9 . 18 eV From the definition of B → given by the expression F → B = q v → × B → , we see that the magnetic force F → B is always perpendicular to v → and B → . (c) The momentum of the proton is p = m v = ( 1 . 67 × 10 - 27 kg ) ( 4 . 00 × 10 4 m s - 1 ) = 7 . 01 × 10 - 23 kg m s - 1

A proton moves through a uniform magnetic field given by B → = ( 10 i ^ - 20 j ^ + 25 k ^ ) mT . At time t 1 , the proton has a velocity given by v → = v x i ^ + v y j ^ + ( 2 . 0 km s - 1 ) k ^ and the magnetic force on the proton is F → B = 4 . 0 × 10 - 17 N i ^ + 2 . 0 × 10 - 17 N j ^ . At that instant, what are (a) v x (b) v y ?

In this question we have been given that a proton is traveling in a given magnetic field with velocity due to which it experiences a magnetic force, and we have to calculate the x & y velocity components of proton. Given that, magnetic field is, B → = 10 i ^ - 20 j ^ + 25 k ^ mT ⇒ B → = 10 i ^ - 20 j ^ + 25 k ^ × 10 - 3 T , Velocity of the proton is, v → = v x i ^ + v y j ^ + ( 2 . 0 km s - 1 ) k ^ ⇒ v → = v x i ^ + v y j ^ + ( 2000 m s - 1 ) k ^ , Proton charge is, q = 1 . 6 × 10 - 19 C , and Force on the proton is, F → B = 4 i ^ + 2 j ^ × 10 - 17 N . When the charge is moving in a magnetic filed then it experiences the magnetic force given by, F B → = q v → × B → . By substituting the values in the formula, we get, 4 i ^ + 2 j ^ × 10 - 17 = 1 . 6 × 10 - 19 v x i ^ + v y j ^ + 2000 m s - 1 k ^ × 10 i ^ - 20 j ^ + 25 k ^ × 10 - 3 T 4 i ^ + 2 j ^ × 10 - 17 = 1 . 6 × 10 - 22 v x i ^ + v y j ^ + 2000 k ^ × 10 i ^ - 20 j ^ + 25 k ^ Using the identities of cross product, i ^ × j ^ = k ^ , j ^ × k ^ = i ^ & k ^ × i ^ = j ^ we get, 4 i ^ + 2 j ^ × 10 5 = 1 . 6 - 20 v x k ^ - 25 v x j ^ - 10 v y k ^ + 25 v y i ^ + 20000 j ^ + 50000 i ^ 4 i ^ + 2 j ^ × 10 5 = 1 . 6 25 v y + 50000 i ^ + - 25 v x + 20000 j ^ + - 20 v x - 10 v y k ^ By comparing coefficients of i ^ , j ^ & k ^ on both sides we get, Comparing coefficients of k ^ we get, - 20 v x - 10 v y = 0 ⇒ v x = - v y 2 ⋯ 1 Comparing coefficients of i ^ we get, 400000 = 1 . 6 25 v y + 50000 ⇒ 250000 = 25 v y + 50000 v y = 8000 m s - 1 ⇒ v y = 8 × 10 3 m s - 1 = 8 km s - 1 From equation 1 , we can write, v x = - 8 × 10 3 2 ⇒ v x = - 4 × 10 3 m s - 1 = - 4 km s - 1

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Earn 100

A proton traveling at $42.0 °$ with respect to the direction of the magnetic field of strength $2.60 mT$ experiences a magnetic force of $1.17 \times 10^{- 17} N .$ Calculate (a) the proton's speed, (b) its kinetic energy in electron-volts, and (c) its momentum. Take, mass of proton = $1.67 \times 10^{- 27} kg$ and $\sin (42 °) = 0.669$ .

Important Questions on Magnetic Fields

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Principles Of Physics International Student Version>Chapter 28 - Magnetic Fields>Problems>Q 52

In the figure given below, a particle moves along a circle in a region of uniform magnetic field with a magnitude of $B = 5.00 mT$ . The particle is either a proton or an electron (you must decide what it is). It experiences a magnetic force of magnitude $3.20 \times 10^{- 15} N .$ What is (a) the particle's speed (b) the radius of the circle, and (c) the period of the motion?

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Principles Of Physics International Student Version>Chapter 28 - Magnetic Fields>Problems>Q 53

An electron of kinetic energy $600 eV$ circles in a plane perpendicular to a uniform magnetic field. The orbit's radius is $12.5 cm .$ Find

(a) the electron's speed,

(b) the magnetic field magnitude,
(c) the circling frequency, and
(d) the period of the motion.
(e) Through what potential difference will the electron have to be accelerated from rest to reach this kinetic energy?

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Principles Of Physics International Student Version>Chapter 28 - Magnetic Fields>Problems>Q 54

In a certain cyclotron, a proton moves in a circle of radius $0.500 m .$ The magnitude of the magnetic field is $1.00 T$ .

(a) What is the oscillator frequency?

(b) What is the kinetic energy of the proton, in electron-volts?

Take the mass of proton = $1.67 \times 10^{- 27} kg$

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Principles Of Physics International Student Version>Chapter 28 - Magnetic Fields>Problems>Q 55

Estimate the total path length travelled by a deuteron in a cyclotron of radius $53 cm$ and operating frequency $12 MHz$ during the (entire) acceleration process. Assume that the accelerating potential between the dees is $120 kV$ . The magnitude of the magnetic field is $1.0 T$ .

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Principles Of Physics International Student Version>Chapter 28 - Magnetic Fields>Problems>Q 56

A proton moves through a uniform magnetic field given by

\vec{B} = (10 \hat{i} - 20 \hat{j} + 25 \hat{k}) mT .

At time

t_{1},

the proton has a velocity given by

\vec{v} = v_{x} \hat{i} + v_{y} \hat{j} + (2.0 km s^{- 1}) \hat{k}

and the magnetic force on the proton is

{\vec{F}}_{B} = (4.0 \times 10^{- 17} N) \hat{i} + (2.0 \times 10^{- 17} N) \hat{j} .

At that instant, what are (a)

v_{x}

(b)

v_{y} ?

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Principles Of Physics International Student Version>Chapter 28 - Magnetic Fields>Problems>Q 57

A single-turn current loop, carrying a current of

8.00 A,

is in the shape of a right triangle with sides

50 cm, 120 cm

and

130 cm .

The loop is in a uniform magnetic field of magnitude

75.0 mT

whose direction is parallel to the current in the

130 cm

side of the loop. What is the magnitude of the magnetic force on (a) the

130 cm

side, (b) the

50 cm

side and (c) the

120 cm

side?
(d) What is the magnitude of the net force on the loop?

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Principles Of Physics International Student Version>Chapter 28 - Magnetic Fields>Problems>Q 58

In figure (a), two concentric coils, lying in the same plane, carry currents in opposite directions. The current in the larger coil $1$ is fixed. Current $i_{2}$ in coil $2$ can be varied. Figure (b) gives the net magnetic moment of the two-coil system as a function of $i_{2}$ . The vertical axis scale is set by, $μ_{nets} = 2.0 \times 10^{- 5} A m^{2}$ and the horizontal axis scale is set by, $i_{2 s} = 20.0 mA$ . If the current in coil $2$ is then reversed, what is the magnitude of the net magnetic moment of the two-coil system when $i_{2} = 7.0 mA ?$

$(a)$ Question Image

$(b)$ Question Image

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Principles Of Physics International Student Version>Chapter 28 - Magnetic Fields>Problems>Q 59

A mass spectrometer (fig.) is used to separate uranium ions of mass $3.92 \times 10^{- 25} kg$ and charge $3.20 \times 10^{- 19} C$ from related species. The ions are accelerated through a potential difference of $180 kV$ and then pass into a uniform magnetic field, where they are bent in a path of radius $1.00 m$ . After traveling through $180 °$ and passing through a slit of width $1.00 mm$ and height $1.00 cm,$ they are collected in a cup. (a) What is the magnitude of the (perpendicular) magnetic field in the separator? If the machine is used to separate out $100 mg$ of material per hour, calculate (b) the current of the desired ions in the machine and (c) the thermal energy produced in the cup in $1.00 h$ .

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Important Questions on Magnetic Fields

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