A uniform magnetic field B = 1.2 mT is directed vertically upward throughout the volume of a laboratory chamber. A proton (mp =1.67 × 10-27 kg) enters the laboratory horizontally from south to north. Calculate the magnitude of centripetal acceleration of the proton if its speed is 3 × 107 m/s .

Question

A uniform magnetic field B = 1.2 mT is directed vertically upward throughout the volume of a laboratory chamber. A proton (mp =1.67 × 10-27 kg) enters the laboratory horizontally from south to north. Calculate the magnitude of centripetal acceleration of the proton if its speed is 3 × 107 m/s . (A)  3.45 × 1012 m/s2  (B)  1.67 × 1012 m/s2  (C)  5.25 × 1012 m/s2  (D)  2.75 × 1012 m/s2

Tardigrade Admin · Accepted Answer

Centripetal acceleration ac=(Fm/mp) =(Bqv/mp) =(1.2×10-3×1.6×10-19×3×107/1.67×10-27) =3.6×1012 m/ s

$3.45 \times 1 0^{12} m / s^{2}$

$1.67 \times 1 0^{12} m / s^{2}$

$5.25 \times 1 0^{12} m / s^{2}$

$2.75 \times 1 0^{12} m / s^{2}$

Centripetal acceleration $a_{c} = \frac{F _{m}}{m _{p}}$
$= \frac{Bq v}{m _{p}}$
$= \frac{1.2 \times 1 0 ^{- 3} \times 1.6 \times 1 0 ^{- 19} \times 3 \times 1 0 ^{7}}{1.67 \times 1 0 ^{- 27}}$
$= 3.6 \times 1 0^{12} m / s$

Q. A uniform magnetic field B=1.2mT is directed vertically upward throughout the volume of a laboratory chamber. A proton (mp​=1.67×10−27kg) enters the laboratory horizontally from south to north. Calculate the magnitude of centripetal acceleration of the proton if its speed is 3×107m/s .

3.45×1012m/s2

1.67×1012m/s2

5.25×1012m/s2

2.75×1012m/s2

Centripetal acceleration ac​=mp​Fm​​ =mp​Bqv​ =1.67×10−271.2×10−3×1.6×10−19×3×107​ =3.6×1012m/s

$3.45 \times 1 0^{12} m / s^{2}$

$1.67 \times 1 0^{12} m / s^{2}$

$5.25 \times 1 0^{12} m / s^{2}$

$2.75 \times 1 0^{12} m / s^{2}$

Centripetal acceleration $a_{c} = \frac{F _{m}}{m _{p}}$
$= \frac{Bq v}{m _{p}}$
$= \frac{1.2 \times 1 0 ^{- 3} \times 1.6 \times 1 0 ^{- 19} \times 3 \times 1 0 ^{7}}{1.67 \times 1 0 ^{- 27}}$
$= 3.6 \times 1 0^{12} m / s$