Magnetics Ques 42
- A particle having the same charge as of electron moves in a circular path of radius $0.5$ $cm$ under the influence of a magnetic field of $0.5$ $T$. If an electric field of $100$ $V / m$ makes it to move in a straight path, then the mass of the particle is (Take, charge of electron $=1.6 \times 10^{-19} C$ )
(a) $1.6 \times 10^{-19} kg$
(b) $1.6 \times 10^{-27} kg$
(c) $9.1 \times 10^{-31} kg$
(d) $2.0 \times 10^{-24} kg$
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Answer:
Correct Answer: 42.(d)
Solution:
Formula:
Magnetic Force Acting On A Moving Point Charge:
- According to given condition, when a particle having charge same as electron move in a magnetic field on circular path, then the force always acts towards the centre and perpendicular to the velocity.
Here, $\quad B=0.5 T$
$ R=\text { radius of circular path }=5 cm $
Now, the magnetic force is
$ \begin{aligned} & F _m=q(v \times B)=q v B \sin 90^{\circ} \\ & F _m=q v B \quad …….(i) \end{aligned} $
When the electric field applied, then the particle moves in a straight path, then this is the case of velocity selector.
Here, the electric force on charge,
$ F _e=q E \quad …….(ii) $
In velocity selector, $F _m=F _e$
$ \Rightarrow \quad q v B=q E \quad …….(iii) $
Initially particle moves under the magnetic field, So the radius of circular path taken by the particle is
$ R=\frac{m v}{q B} \quad …….(iv) $
From Eqs. (iii) and (iv),
$ \begin{aligned} & m=\frac{q B^{2} R}{E} \\ & m=\frac{1.6 \times 10^{-19} \times 0.25 \times 0.5 \times 10^{-2}}{10^{2}} \\ & m=2 \times 10^{-24} kg \end{aligned} $
BETA
