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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 $1.6 \times 10^{-19} \mathrm{C}$ )
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Verified Answer
The correct answer is:
$2.4 \times 10^{-12} \mathrm{~N}$
Here: $q=1.6 \times 10^{-19} \mathrm{C}, B=1.5 \mathrm{~Wb} / \mathrm{m}^2$,
$v=2 \times 10^7 \mathrm{~m} / \mathrm{s}, \theta=30^{\circ}$ or $\sin 30^{\circ}=\frac{1}{2}$
Force on proton is given by
$F=q v B \sin \theta$
$\begin{aligned} & =1.6 \times 10^{-19} \times 2 \times 10^7 \times 1.5 \times \frac{1}{2} \\ & =2.4 \times 10^{-12} \mathrm{~N}\end{aligned}$
$v=2 \times 10^7 \mathrm{~m} / \mathrm{s}, \theta=30^{\circ}$ or $\sin 30^{\circ}=\frac{1}{2}$
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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