One proton beam enters a magnetic field of $10^{- 4}$ T normally, Specific charge $10^{11} C / k g$ . Velocity = $10^{7}$ m/s What is the radius of the circle described by it?

0.1 m

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10 m

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1 m

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Solution

The correct option is C $1 m$
The radius of the proton in the circular path is given by:
$r = \frac{m v}{q B}$

$= \frac{10^{7}}{10^{11} \times 10^{- 4}}$

$= 1 m (∵ q / m = 10^{11} C / k g)$

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One proton beam enters a magnetic field of 10−4 T normally, Specific charge 1011C/kg. Velocity = 107 m/s What is the radius of the circle described by it?

The correct option is C 1 mThe radius of the proton in the circular path is given by:r=mvqB=1071011×10−4=1m(∵q/m=1011C/kg)

One proton beam enters a magnetic field of $10^{- 4}$ T normally, Specific charge $10^{11} C / k g$ . Velocity = $10^{7}$ m/s What is the radius of the circle described by it?

The correct option is C $1 m$
The radius of the proton in the circular path is given by:
$r = \frac{m v}{q B}$

$= \frac{10^{7}}{10^{11} \times 10^{- 4}}$

$= 1 m (∵ q / m = 10^{11} C / k g)$