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Q.

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)

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a

0.71mT

b

0.071mT

c

71mT

d

7.1mT

answer is A.

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Detailed Solution

∵K.E of proton=12mV2 ⇒1×106×1.6×10−19=12×1.6×10−27×V2⇒V=2×107 m/s∴   Magnetic Force accelerates the proton ∴BqV= ma  ⇒B=maqV  ⇒B=1.6×10−27×10121.6×10−19×2×107                   =0.71×10−3TSo,  B = 0.71mT
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