Solved Numerical Problem No.4 ( Unit-13: Electromagnetism ); HSSC-II (Class-12); Physics (New Book/KPK Text Board, Peshawar) With Verified Answers

An 8 MeV proton enters perpendicularly into a uniform magnetic field of 2.5 T. (a) Find the force on the electron (b) What will be the radius of the path of the proton? (Answer: 1.6 ｘ 10⁻¹¹ N, 0.17 m)

Given:

K.E. of Proton = 8 MeV = 8 ｘ 10⁶ ｘ 1.6 ｘ 10⁻¹⁹ J = 1.28 ｘ 10⁻¹² J

Angle = θ = 90°

Magnetic Field = B = 2.5 T

Force = F = 2 ｘ 10⁻¹² N

[ ∴ Charge of proton = q = 1.6 ｘ 10⁻¹⁹ C]
[ ∴ Mass of proton = m = 1.67 ｘ 10⁻²⁷ kg]

To Find:

(a) Magnetic Force on Electron = F = ?

(b) Radius of the path of Proton = r = ?

Solution:

(a) Magnetic Force on Electron = F = ?

Since we have the magnetic force formula:

F = q ( v ｘ B )

F = q v B sinθ --------------------(1)

To find the magnitude of the magnetic force, we have to first find the velocity of the Proton. Keeping the given quantity we have a formula for K.E.

K.E. =

$\frac {1}{2}$ mv²

or

v² =

$\frac {2 K.E.}{m}$

Or

v =

$\sqrt frac {2 K.E.}{m}$

by putting the corresponding values

v =

$\sqrt frac {2 ｘ 1.28 ｘ 10⁻¹² J}{1.67 ｘ 10⁻²⁷ kg}$

v =

$\sqrt frac {2.56 ｘ 10⁻¹² J}{1.67 ｘ 10⁻²⁷ kg}$

v =

$\sqrt {1.533 ｘ 10⁻¹²⁺²⁷ m² s⁻²}$

v =

$\sqrt {1.533 ｘ 10 ¹⁵ m² s⁻²}$

or

v =

$\sqrt {15.33 ｘ 10¹⁴ m² s⁻²}$

v = 3.915 ｘ 10⁷ m s⁻¹ --------------(2)

Now putting the value of velocity v of proton eqn (2) and other corresponding values in equation (1)

F = q v B sinθ

F = 1.6 ｘ 10⁻¹⁹ J ｘ 3.915 ｘ 10⁷ m s⁻¹ ｘ 2.5 T sin90°

{ sin90° = 1}

F = 15.666 ｘ 10⁻¹⁹⁺⁷ N

F = 15.666 ｘ 10⁻¹² N

or

F = 1.5666 ｘ 10⁻¹¹ N ------------------ Ans. (1)

-----------------------------------------------

(b) Radius of the path of Proton = r = ?

We have a formula for the charge (e ) to mass (m) ratio:

$\frac {e}{m}$ =

$\frac {v}{Br}$

or

r =

$\frac {vm}{eB}$

by putting values

r =

$\frac {3.915 ｘ 10⁷ m s⁻¹ ｘ 1.67 ｘ 10⁻²⁷ kg }{1.6 ｘ 10⁻¹⁹ C ｘ 2.5 T}$

r =

$\frac {6.538ｘ 10⁻²⁰ kg m s⁻¹}{4 ｘ 10⁻¹⁹ C T}$

r = 1.634 ｘ 10⁻¹ m

or

r = 0.1634 m -----------------------Ans. (2)

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