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

Question

Proton with kinetic energy of 1 MeV moves from south to north It gets an acceleration of 1012 m/s2 by an applied magnetic field (west to east). The value of magnetic field: (Rest mass of proton is 1.6 × 10-27 kg) (A) 7.1 mT (B) 71 mT (C) 0.071 mT (D) 0.71 mT

Tardigrade Admin · Accepted Answer

a=(qvB/m) B=(ma/qv)=(ma√m/√2k) =(m3/2a/e√2k)=((1.6×10-27)3/2×1012/1.6×10-19√2×1×106×1.6×10-19) =0.71 m T

Q. Proton with kinetic energy of $1 M e V$ moves from south to north It gets an acceleration of $1 0^{12} m / s^{2}$ by an applied magnetic field (west to east). The value of magnetic field : (Rest mass of proton is $1.6 \times 1 0^{- 27} k g$ )

7.1 mT

71 mT

0.071 mT

0.71 mT

$a = \frac{q v B}{m}$
$B = \frac{ma}{q v} = \frac{ma m}{2 k}$
$= \frac{m ^{3/2} a}{e 2 k} = \frac{( 1.6 \times 1 0 ^{- 27} ) ^{3/2} \times 1 0 ^{12}}{1.6 \times 1 0 ^{- 19} 2 \times 1 \times 1 0 ^{6} \times 1.6 \times 1 0 ^{- 19}}$
$= 0.71 m T$

Q. Proton with kinetic energy of 1MeV 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)