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A particle of mass $2.2 \times 10^{-30} \mathrm{~kg}$ and charge $1.6 \times 10^{-19} \mathrm{C}$ is moving at a speed of $10 \mathrm{~km} \mathrm{~s}^{-1}$ in a circular path of radius $2.8 \mathrm{~cm}$ inside a solenoid. The solenoid has $25 \frac{\text { turns }}{\mathrm{cm}}$ and its magnetic field is perpendicular to the plane of the particle's path. The current in the solenoid is
$\left(\right.$ Take, $\mu_0=4 \pi \times 10^{-7} \mathrm{Hm}^{-1}$ )
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$\left(\right.$ Take, $\mu_0=4 \pi \times 10^{-7} \mathrm{Hm}^{-1}$ )
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Verified Answer
The correct answer is:
$1.56 \mathrm{~mA}$
Given, mass of the particle,
$m=22 \times 10^{-30} \mathrm{~kg}$
Charge, $q=1.6 \times 10^{-19} \mathrm{C}$
Velocity, $v=10 \mathrm{kms}^{-1}=10^4 \mathrm{~ms}^{-1}$
Radius of circular path, $r=2.8 \mathrm{~cm}=2.8 \times 10^{-2} \mathrm{~m}$
$n=25 \mathrm{turns} / \mathrm{cm}=2500 \frac{\mathrm{turns}}{\mathrm{m}}$
Since, $r=m v / B q$ ...(i)
Magnetic field inside the solenoid,
$B=\mu_0 n I$ ...(ii)
From Eqs. (i) and (ii), we get
$r=\frac{m v}{\left(\mu_0 n I\right) q} \Rightarrow I=\frac{m v}{\mu_0 n q r}$
$\begin{aligned} & =\frac{2.2 \times 10^{-30} \times 10^4}{4 \pi \times 10^{-7} \times 2500 \times 1.6 \times 10^{-19} \times 2.8 \times 10^{-2}} \\ & =1.56 \times 10^{-3} \mathrm{~A}=1.56 \mathrm{~mA}\end{aligned}$
$m=22 \times 10^{-30} \mathrm{~kg}$
Charge, $q=1.6 \times 10^{-19} \mathrm{C}$
Velocity, $v=10 \mathrm{kms}^{-1}=10^4 \mathrm{~ms}^{-1}$
Radius of circular path, $r=2.8 \mathrm{~cm}=2.8 \times 10^{-2} \mathrm{~m}$
$n=25 \mathrm{turns} / \mathrm{cm}=2500 \frac{\mathrm{turns}}{\mathrm{m}}$
Since, $r=m v / B q$ ...(i)
Magnetic field inside the solenoid,
$B=\mu_0 n I$ ...(ii)
From Eqs. (i) and (ii), we get
$r=\frac{m v}{\left(\mu_0 n I\right) q} \Rightarrow I=\frac{m v}{\mu_0 n q r}$
$\begin{aligned} & =\frac{2.2 \times 10^{-30} \times 10^4}{4 \pi \times 10^{-7} \times 2500 \times 1.6 \times 10^{-19} \times 2.8 \times 10^{-2}} \\ & =1.56 \times 10^{-3} \mathrm{~A}=1.56 \mathrm{~mA}\end{aligned}$
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