A proton ( m p =1.67 × 10-27 kg . and . q p =1.6 × 10-19 C ) enters perpendicular to the magnetic field of intensity 2 Wb / m 2 with a velocity 3.4 × 107 m / sec. The acceleration of the proton should be

Question

A proton ( m p =1.67 × 10-27 kg . and . q p =1.6 × 10-19 C ) enters perpendicular to the magnetic field of intensity 2 Wb / m 2 with a velocity 3.4 × 107 m / sec. The acceleration of the proton should be (A) 6.5 × 1015 m / sec 2 (B) 6.5 × 1013 m / sec 2 (C) 6.5 × 1011 m / sec 2 (D) 6.5 × 109 m / sec 2

Tardigrade Admin · Accepted Answer

As we know that F = ma qvB sin 90°= ma a =( qvB / m ) a=(1.6 × 10-19 × 2 × 3.4 × 107/1.67 × 10-27) a=6.5 × 1015 m / sec

Q. A proton $(m_{p} = 1.67 \times 1 0^{- 27} k g$ and $q_{p} = 1.6 \times 1 0^{- 19} C)$ enters perpendicular to the magnetic field of intensity $2 Wb / m^{2}$ with a velocity $3.4 \times 1 0^{7} m / sec$ . The acceleration of the proton should be

$6.5 \times 1 0^{15} m / se c^{2}$

$6.5 \times 1 0^{13} m / se c^{2}$

$6.5 \times 1 0^{11} m / se c^{2}$

$6.5 \times 1 0^{9} m / se c^{2}$

As we know that
$F = ma$
$q v B sin 9 0^{\circ} = ma$
$a = \frac{q v B}{m}$
$a = \frac{1.6 \times 1 0 ^{- 19} \times 2 \times 3.4 \times 1 0 ^{7}}{1.67 \times 1 0 ^{- 27}}$
$a = 6.5 \times 1 0^{15} m / sec$

Q. A proton (mp​=1.67×10−27kg and qp​=1.6×10−19C) enters perpendicular to the magnetic field of intensity 2Wb/m2 with a velocity 3.4×107m/sec. The acceleration of the proton should be

6.5×1015m/sec2

6.5×1013m/sec2

6.5×1011m/sec2

6.5×109m/sec2