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Physics Ncert Solutions Class 12th Physics Class 12th Magnetism and Matter

5.22 A monoenergetic (18 keV) electron beam initially in the horizontal direction is subjected to a horizontal magnetic field of 0.04 G normal to the initial direction. Estimate the up or down deflection of the beam over a distance of 30 cm (me = 9.11 × 10^–31kg).

[Note: Data in this exercise are so chosen that the answer will give you an idea of the effect of earth’s magnetic field on the motion of the electron beam from the electron gun to the screen in a TV set.]

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Payal Gupta | Contributor-Level 10

4 months ago

5.22 Energy of an electron beam, E = 18 keV = 18 $\times 10^{3}$ eV

Charge of an electron, e = 1.6 $\times$ $10^{- 19}$ C

Total energy of the electron beam, $E_{T}$ = E $\times e$ = 18 $\times 10^{- 3}$ $\times$ 1.6 $\times$ $10^{- 19}$

Magnetic field, B = 0.04 G

Mass of an electron, $m_{e}$ = 9.11 $\times 10^{- 31}$ kg

Distance up to which the electron beam travels, d = 30 cm = 0.3 m

The kinetic energy of an electron beam, $E_{k}$ = $\frac{1}{2}$ m $v^{2}$ = $E_{T}$

v = $\sqrt{\frac{2 E_{T}}{m}}$ = $\sqrt{\frac{2 \times 18 \times 10^{3} \times 1.6 \times 10^{- 19}}{9.11 \times 10^{- 31}}}$ = 79.51 $\times 10^{6}$ m/s

The electron beam deflects along a circular path of radius r.

The force due to the magnetic field balances the centripetal force of the path

Bev = $\frac{m v^{2}}{r}$ or r = $\frac{m v}{B e}$ = $\frac{9.11 \times 10^{- 31} \times 79.51 \times 10^{6}}{0.4 \times 10^{- 4} \times 1.6 \times 10^{- 19}}$ = 11.3 m

Let the up and down deflection of the electron beam be x = r (1- cos) where $θ$ = Angle of declination

$\sin ? θ$ = $\frac{d}{r}$ = $\frac{0.3}{11.3}$

$θ = 1.521 °$

x = 11.3 (1 &nd

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5.22 Energy of an electron beam, E = 18 keV = 18 $\times 10^{3}$ eV

Charge of an electron, e = 1.6 $\times$ $10^{- 19}$ C

Total energy of the electron beam, $E_{T}$ = E $\times e$ = 18 $\times 10^{- 3}$ $\times$ 1.6 $\times$ $10^{- 19}$

Magnetic field, B = 0.04 G

Mass of an electron, $m_{e}$ = 9.11 $\times 10^{- 31}$ kg

Distance up to which the electron beam travels, d = 30 cm = 0.3 m

The kinetic energy of an electron beam, $E_{k}$ = $\frac{1}{2}$ m $v^{2}$ = $E_{T}$

v = $\sqrt{\frac{2 E_{T}}{m}}$ = $\sqrt{\frac{2 \times 18 \times 10^{3} \times 1.6 \times 10^{- 19}}{9.11 \times 10^{- 31}}}$ = 79.51 $\times 10^{6}$ m/s

The electron beam deflects along a circular path of radius r.

The force due to the magnetic field balances the centripetal force of the path

Bev = $\frac{m v^{2}}{r}$ or r = $\frac{m v}{B e}$ = $\frac{9.11 \times 10^{- 31} \times 79.51 \times 10^{6}}{0.4 \times 10^{- 4} \times 1.6 \times 10^{- 19}}$ = 11.3 m

Let the up and down deflection of the electron beam be x = r (1- cos) where $θ$ = Angle of declination

$\sin ? θ$ = $\frac{d}{r}$ = $\frac{0.3}{11.3}$

$θ = 1.521 °$

x = 11.3 (1 – cos 1.521 $°$ ) = 3.98 mm

Therefore up and down deflection of the beam is 3.98 mm.

less

5.22 Energy of an electron beam, E = 18 keV = 18 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msup> </math> eVCharge of an electron, e = 1.6 <math> <mo>×</mo> </math> <math> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>19</mn> </mrow> </mrow> </msup> </math> CTotal energy of the electron beam, <math> <msub> <mrow> <mrow> <mi>E</mi> </mrow> </mrow> <mrow> <mrow> <mi>T</mi> </mrow> </mrow> </msub> </math> = E <math> <mo>×</mo> <mi>e</mi> </math> = 18 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>3</mn> </mrow> </mrow> </msup> </math> <math> <mo>×</mo> </math> 1.6 <math> <mo>×</mo> </math> <math> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>19</mn> </mrow> </mrow> </msup> </math> Magnetic field, B = 0.04 GMass of an electron, <math> <msub> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> <mrow> <mrow> <mi>e</mi> </mrow> </mrow> </msub> </math> = 9.11 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>31</mn> </mrow> </mrow> </msup> </math> kgDistance up to which the electron beam travels, d = 30 cm = 0.3 mThe kinetic energy of an electron beam, <math> <msub> <mrow> <mrow> <mi>E</mi> </mrow> </mrow> <mrow> <mrow> <mi>k</mi> </mrow> </mrow> </msub> </math> = <math> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> </math> m <math> <msup> <mrow> <mrow> <mi>v</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </math> = <math> <msub> <mrow> <mrow> <mi>E</mi> </mrow> </mrow> <mrow> <mrow> <mi>T</mi> </mrow> </mrow> </msub> </math> v = <math> <msqrt> <mrow> <mfrac> <mrow> <mrow> <mn>2</mn> <msub> <mrow> <mrow> <mi>E</mi> </mrow> </mrow> <mrow> <mrow> <mi>T</mi> </mrow> </mrow> </msub> </mrow> </mrow> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> </mfrac> </mrow> </msqrt> </math> = <math> <msqrt> <mrow> <mfrac> <mrow> <mrow> <mn>2</mn> <mo>×</mo> <mn>18</mn> <mi mathvariant="normal"> </mi> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msup> <mi> </mi> <mo>×</mo> <mi> </mi> <mn>1.6</mn> <mo>×</mo> <mi mathvariant="normal"> </mi> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>19</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mn>9.11</mn> <mi mathvariant="normal"> </mi> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>31</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> </mrow> </msqrt> </math> = 79.51 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>6</mn> </mrow> </mrow> </msup> </math> m/sThe electron beam deflects along a circular path of radius r.The force due to the magnetic field balances the centripetal force of the pathBev = <math> <mfrac> <mrow> <mrow> <mi>m</mi> <msup> <mrow> <mrow> <mi>v</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mi>r</mi> </mrow> </mrow> </mfrac> </math> or r = <math> <mfrac> <mrow> <mrow> <mi>m</mi> <mi>v</mi> </mrow> </mrow> <mrow> <mrow> <mi>B</mi> <mi>e</mi> </mrow> </mrow> </mfrac> </math> = <math> <mfrac> <mrow> <mrow> <mn>9.11</mn> <mi mathvariant="normal"> </mi> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>31</mn> </mrow> </mrow> </msup> <mo>×</mo> <mn>79.51</mn> <mi mathvariant="normal"> </mi> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>6</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mn>0.4</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>4</mn> </mrow> </mrow> </msup> <mo>×</mo> <mn>1.6</mn> <mo>×</mo> <mi mathvariant="normal"> </mi> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>19</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> </math> = 11.3 mLet the up and down deflection of the electron beam be x = r (1- cos) where <math> <mi>θ</mi> </math> = Angle of declination <math> <mrow> <mrow> <mi mathvariant="normal">sin</mi> </mrow> <mo>?</mo> <mrow> <mi>θ</mi> </mrow> </mrow> </math> = <math> <mfrac> <mrow> <mrow> <mi>d</mi> </mrow> </mrow> <mrow> <mrow> <mi>r</mi> </mrow> </mrow> </mfrac> </math> = <math> <mfrac> <mrow> <mrow> <mn>0.3</mn> </mrow> </mrow> <mrow> <mrow> <mn>11.3</mn> </mrow> </mrow> </mfrac> </math> <math> <mi>θ</mi> <mo>=</mo> <mn>1.521</mn> <mo>°</mo> </math> x = 11.3 (1 – cos 1.521 <math> <mo>°</mo> </math> ) = 3.98 mmTherefore up and down deflection of the beam is 3.98 mm.

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