A proton enters a magnetic field of flux density 1.5 Wb / m 2 with a speed of 2 × 107 m / s at angle of 30° with the field. The force on a proton will be

Question

A proton enters a magnetic field of flux density 1.5 Wb / m 2 with a speed of 2 × 107 m / s at angle of 30° with the field. The force on a proton will be (A) 0.24 × 10-12 N (B) 2.4 × 10-12 N (C) 24 × 10-12 N (D) 0.024 × 10-12 N

Tardigrade Admin · Accepted Answer

Magnetic force, F=q v B sin θ ∴ F=(1.6 × 10-19) ×(2 × 107) ×(1.5) sin 30° F=1.6 × 10-12 × 2 × 1.5 × (1/2) F=2.4 × 10-12 N

Q. A proton enters a magnetic field of flux density $1.5 Wb / m^{2}$ with a speed of $2 \times 1 0^{7} m / s$ at angle of $3 0^{\circ}$ with the field. The force on a proton will be

$0.24 \times 1 0^{- 12} N$

$2.4 \times 1 0^{- 12} N$

$24 \times 1 0^{- 12} N$

$0.024 \times 1 0^{- 12} N$

Magnetic force,
$F = q v B sin θ$
$∴ F = (1.6 \times 1 0^{- 19}) \times (2 \times 1 0^{7}) \times (1.5) sin 3 0^{\circ}$
$F = 1.6 \times 1 0^{- 12} \times 2 \times 1.5 \times \frac{1}{2}$
$F = 2.4 \times 1 0^{- 12} N$

Q. A proton enters a magnetic field of flux density 1.5Wb/m2 with a speed of 2×107m/s at angle of 30∘ with the field. The force on a proton will be

0.24×10−12N

2.4×10−12N

24×10−12N

0.024×10−12N