A proton enters a magnetic field of intensity 1.5 Wb / m 2 with a velocity 2 × 107 m / s in a direction at an angle 30° with the field. The force on the proton will be (charge on proton is 1.6 × 10-19 C )

Question

A proton enters a magnetic field of intensity 1.5 Wb / m 2 with a velocity 2 × 107 m / s in a direction at an angle 30° with the field. The force on the proton will be (charge on proton is 1.6 × 10-19 C ) (A) 2.4 × 10-12 N (B) 4.8 × 10-12 N (C) 1.2 × 10-12 N (D) 7.2 × 10-12 N

Tardigrade Admin · Accepted Answer

Here: q=16 × 10-19 C, B=15 Wb / m 2, v=2 × 107 m / s, θ=30° or sin 30°=(1/2) Force on proton is given by F =q v B sin θ =1.6 × 10-19 × 2 × 107 × 1.5 × (1/2) =2.4 × 10-12 N

Q. A proton enters a magnetic field of intensity $1.5 Wb / m^{2}$ with a velocity $2 \times 1 0^{7} m / s$ in a direction at an angle $3 0^{\circ}$ with the field. The force on the proton will be (charge on proton is $1.6 \times 1 0^{- 19} C$ )

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

$4.8 \times 1 0^{- 12} N$

$1.2 \times 1 0^{- 12} N$

$7.2 \times 1 0^{- 12} N$

Here: $q = 16 \times 1 0^{- 19} C, B = 15 Wb / m^{2}$ ,
$v = 2 \times 1 0^{7} m / s, θ = 3 0^{\circ}$ or $sin 3 0^{\circ} = \frac{1}{2}$
Force on proton is given by
$F = q v B sin θ$
$= 1.6 \times 1 0^{- 19} \times 2 \times 1 0^{7} \times 1.5 \times \frac{1}{2}$
$= 2.4 \times 1 0^{- 12} N$

Q. A proton enters a magnetic field of intensity 1.5Wb/m2 with a velocity 2×107m/s in a direction at an angle 30∘ with the field. The force on the proton will be (charge on proton is 1.6×10−19C )

2.4×10−12N

4.8×10−12N

1.2×10−12N

7.2×10−12N