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Standard XII

Physics

Analysis of Force Equation

A proton beam...

Question

A proton beam enters a magnetic field of $10^{- 4} W b m^{- 2}$ normally. If the specific charge of the proton is $10^{11} C k g^{- 1}$ and its velocity is $10^{9} m s^{- 1}$ , then the radius of the circle described will be

0.1 m

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10 m

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100 m

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1 m

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Solution

The correct option is C $100 m$
Given : $B = 10^{- 4}$ $W b / m^{2}$ $q = 10^{11} C / k g$ $v = 10^{9} m / s$

Radius of circle described, $r = \frac{m v}{B q}$
$∴$ $r = \frac{10^{9}}{10^{- 4} \times 10^{11}} = 100 m$

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A proton beam enters a magnetic field of 10−4Wb m−2 normally. If the specific charge of the proton is 1011C kg−1 and its velocity is 109 ms−1, then the radius of the circle described will be

The correct option is C 100 mGiven : B=10−4 Wb/m2 q=1011C/kg v=109m/sRadius of circle described, r=mvBq∴ r=10910−4×1011=100 m

A proton beam enters a magnetic field of $10^{- 4} W b m^{- 2}$ normally. If the specific charge of the proton is $10^{11} C k g^{- 1}$ and its velocity is $10^{9} m s^{- 1}$ , then the radius of the circle described will be

The correct option is C $100 m$
Given : $B = 10^{- 4}$ $W b / m^{2}$ $q = 10^{11} C / k g$ $v = 10^{9} m / s$

Radius of circle described, $r = \frac{m v}{B q}$
$∴$ $r = \frac{10^{9}}{10^{- 4} \times 10^{11}} = 100 m$