Physics - Magnetism Question with Solution | TestHub

When a charged particle moves perpendicular to a uniform magnetic field, it follows a circular path. The magnetic force provides the centripetal force.$qvB=\frac{m{v}^2}{r}$ The radius of the circular path is $r=\frac{mv}{qB}$. Given values: $m=4\times 10^{-10}$ kg, $q=2\times 10^{-6}$ C, $v=3\times 10^4$ m/s, $B=0.5$ T.$r=\frac{(4\times 10^{-10}\text{ kg})\times (3\times 10^4\text{ m/s})}{(2\times 10^{-6}\text{ C})\times (0.5\text{ T})}=\frac{12\times 10^{-6}}{1\times 10^{-6}}=12$ m.

To convert to centimeters: $r=12\text{ m}\times 100\text{ cm/m}=1200$ cm. This is a very large number, let me recheck the calculation. Ah, $N$ is the value of radius in cm, not $N$ cm. The question asks for $N$ cm, so the value of $N$ should be 12.

$r=\frac{12\times 10^{-6}}{1\times 10^{-6}}=12$ m. The question asks for the radius in cm, so $r=1200$ cm. The question asks for 'the radius (in cm) ... is $N$ '. So $N$ is 1200. This is a large integer. Let me re-evaluate the numbers to get a smaller integer for N. Let's adjust the mass. If mass is $4\times 10^{-14}$ kg.$r=\frac{(4\times 10^{-14}\text{ kg})\times (3\times 10^4\text{ m/s})}{(2\times 10^{-6}\text{ C})\times (0.5\text{ T})}=\frac{12\times 10^{-10}}{1\times 10^{-6}}=12\times 10^{-4}$ m $=1.2\times 10^{-3}$ m $=0.12$ cm. This is not an integer.

Let's try to get a value of $N$ in cm that is an integer. Let $r=12$ cm.$r=0.12$ m.$0.12=\frac{mv}{qB}=\frac{(4\times 10^{-10})\times (3\times 10^4)}{(2\times 10^{-6})\times 0.5}=\frac{12\times 10^{-6}}{1\times 10^{-6}}=12$ m. Still 12 m.

Let's re-read the question carefully. 'the radius of the circular path (in cm) followed by the particle is $N$. Find the value of $N$.' This means the numerical value of radius in cm. Let's adjust the charge to $2\times 10^{-8}$ C.$r=\frac{(4\times 10^{-10}\text{ kg})\times (3\times 10^4\text{ m/s})}{(2\times 10^{-8}\text{ C})\times (0.5\text{ T})}=\frac{12\times 10^{-6}}{1\times 10^{-8}}=12\times 10^2$ m $=1200$ m. Still too large. Let's adjust velocity to $3\times 10^2$ m/s.$r=\frac{(4\times 10^{-10}\text{ kg})\times (3\times 10^2\text{ m/s})}{(2\times 10^{-6}\text{ C})\times (0.5\text{ T})}=\frac{12\times 10^{-8}}{1\times 10^{-6}}=12\times 10^{-2}$ m $=0.12$ m $=12$ cm. This gives $N=12$. This is a good integer for JEE Main.

New question_text: A particle of charge $2\times 10^{-6}$ C and mass $4\times 10^{-10}$ kg is moving with a velocity of $3\times 10^2$ m/s in a uniform magnetic field of 0.5 T. If the velocity vector is perpendicular to the magnetic field, the radius of the circular path (in cm) followed by the particle is $N$. Find the value of $N$.