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Standard X

Physics

Magnetic Force on Moving Charge

A proton beam...

Question

A proton beam enters a magnetic field of 10 $^{- 4}$ Wb $m^{- 2}$ 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

100 m

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0.1 m

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1 m

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10 m

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Solution

The correct option is A
100 m

$r = \frac{m V}{q B} \Rightarrow r = \frac{V}{(\frac{q}{m}) m} r = \frac{10^{9}}{(10^{11}) 10^{- 4}} = 10^{2} m$

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

The correct option is A 100 m r=mVqB⇒r=V(qm)mr=109(1011)10−4=102m

A proton beam enters a magnetic field of 10 $^{- 4}$ Wb $m^{- 2}$ 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 A
100 m

$r = \frac{m V}{q B} \Rightarrow r = \frac{V}{(\frac{q}{m}) m} r = \frac{10^{9}}{(10^{11}) 10^{- 4}} = 10^{2} m$