Q. 123.6( 18 Votes )

# In Exercise 4.11

Answer :

Given:

Magnetic field strength, B = 6.5 G = 6.5 × 10^{-4} T

Initial velocity of electron = 4.8 × 10^{6} ms^{-1}

Angle between the initial velocity of electron and magnetic field, θ = 90^{0}

We can relate the velocity of the electron to its angular frequency by the relation,

V = rω …(1)

Where,

V = velocity of electron

r = radius of path

ω = angular frequency

We understand that, magnetic force on the electron is equal to the centripetal force on it, hence we can write,

F_{e} = F_{c}

e × V × B = mV^{2}/r …(2)

From equation (2) we can write,

⇒ eB = mV/r …(3)

Now, by putting value of V from equation (1) in equation (3)

⇒ e × B = m × ω …(4)

We know that, ω = 2πν

Putting in equation (4), we have

…(5)

Now, by putting the values in equation (5) we get,

⇒

⇒ v = 18.2 × 10^{6}Hz

Hence, the frequency of rotation is 18.2 × 10^{6}Hz.

Note: The frequency of rotation is independent of the initial velocity of the electron.

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