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Q.

Proton with kinetic energy of 1 MeV moves from south to north. It gets an acceleration of $10^{12} m / s^{2}$ by an applied magnetic field (west to east). The value of magnetic field: (Rest mass of proton is $1.6 \times 10^{- 27} k g)$

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a

$0.071 m T$

b

$7.1 m T$

c

$0.71 m T$

d

$71 m T$

answer is A.

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Detailed Solution

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$∵ K . E of proton = \frac{1}{2} {mV}^{2} \Rightarrow 1 \times 10^{6} \times 1 .6 \times 10^{- 19} = \frac{1}{2} \times 1 .6 \times 10^{- 27} \times V^{2}$

$\Rightarrow V = \sqrt{2} \times 10^{7} m / s$

$∴$ $Magnetic Force accelerates the proton ∴ BqV = ma \Rightarrow B = \frac{ma}{qV} \Rightarrow B = \frac{1 .6 \times 10^{- 27} \times 10^{12}}{1 .6 \times 10^{- 19} \times \sqrt{2} \times 10^{7}}$

$= 0.71 \times 10^{- 3} T$

So, B = 0.71mT

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Proton with kinetic energy of 1 MeV moves from south to north. It gets an acceleration of 1012m/s2 by an applied magnetic field (west to east). The value of magnetic field: (Rest mass of proton is 1.6×10−27kg)