A proton enters a magnetic field of intensity 1.5 ~Wb / m^2 with a velocity 2 × 10^7 ~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 )

Correct answer: A. 2.4 × 10^ -12 ~N Here: q=1.6 × 10^ -19 C, B=1.5 ~Wb / m^2 , v=2 × 10^7 ~m / s, θ=30^ or 30^ =(1)/(2) Force on proton is given by F=q v B θ aligned & =1.6 × 10^ -19 × 2 × 10^7 × 1.5 × (1)/(2) & =2.4 × 10^ -12 ~N aligned

JIPMER 2004

Exam: JIPMER
Year: 2004
Subject: Physics
Chapter: Magnetic Effects of Current

JIPMER 2004 · Physics · Magnetic Effects of Current

A proton enters a magnetic field of intensity 1.5 ~Wb / m^2 with a...

A proton enters a magnetic field of intensity

1.5 \mathrm{~Wb} / \mathrm{m}^2

with a velocity

2 \times 10^7 \mathrm{~m} / \mathrm{s}

in a direction at an angle

30^{\circ}

with the field. The force on the proton will be (charge on proton is

1.6 \times 10^{-19} \mathrm{C}

)

Options

A $2.4 \times 10^{-12} \mathrm{~N}$
B $4.8 \times 10^{-12} \mathrm{~N}$
C $1.2 \times 10^{-12} \mathrm{~N}$
D $7.2 \times 10^{-12} \mathrm{~N}$

Correct answer

A. $2.4 \times 10^{-12} \mathrm{~N}$

Step-by-step solution

Here:

q=1.6 \times 10^{-19} \mathrm{C}, B=1.5 \mathrm{~Wb} / \mathrm{m}^2

v=2 \times 10^7 \mathrm{~m} / \mathrm{s}, \theta=30^{\circ}

\sin 30^{\circ}=\frac{1}{2}

Force on proton is given by

F=q v B \sin \theta

\begin{aligned} & =1.6 \times 10^{-19} \times 2 \times 10^7 \times 1.5 \times \frac{1}{2} \\ & =2.4 \times 10^{-12} \mathrm{~N}\end{aligned}

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