Question

A proton moving with a velocity of $8\times 10^5$ ms$^{-1}$ enters a uniform magnetic field normal to the direction of the magnetic field. If the radius of the circular path of the proton in the magnetic field is $8.3$ cm, then the magnitude of the magnetic field is (Charge of proton = $1.6\times 10^{-19}$ C and mass of the proton = $1.66\times 10^{-27}$ kg)

• A. 500 mT
• B. 100 mT
• C. 200 mT
• D. 400 mT

Hint:

For a charged particle moving perpendicular to a magnetic field, the radius of the circular path is given by $R = mv/qB$. It is crucial to perform calculations using SI units (meters, kilograms, Coulombs, Teslas) and only convert to millitesla (mT) at the very end.

Answer 1

The Correct Option is B

Step 1: State the formula for the radius of a charged particle's path in a magnetic field.
When a charged particle enters a uniform magnetic field perpendicularly ($\theta=90^\circ$), the magnetic force provides the centripetal force, and the particle moves in a circular path. The radius $R$ of the path is given by: \[ R = \frac{mv}{qB}. \] where $m$ is the mass, $v$ is the velocity, $q$ is the charge, and $B$ is the magnetic field magnitude.

Step 2: Solve the formula for the magnetic field $B$.
Rearrange the formula to isolate $B$: \[ B = \frac{mv}{qR}. \]

Step 3: Convert the given values to standard units (SI).
Mass of proton: $m = 1.66 \times 10^{-27}$ kg. Velocity: $v = 8 \times 10^5$ m/s. Charge of proton: $q = 1.6 \times 10^{-19}$ C. Radius of path: $R = 8.3 \text{ cm} = 8.3 \times 10^{-2} \text{ m}$.

Step 4: Substitute the numerical values and calculate B.
\[ B = \frac{(1.66 \times 10^{-27} \text{ kg}) (8 \times 10^5 \text{ m/s})}{(1.6 \times 10^{-19} \text{ C}) (8.3 \times 10^{-2} \text{ m})}. \] Group the numerical and exponential parts: \[ B = \frac{1.66 \times 8}{1.6 \times 8.3} \times \frac{10^{-27} \times 10^5}{10^{-19} \times 10^{-2}} \text{ T}. \] Simplify the numerical part: \[ \frac{1.66 \times 8}{1.6 \times 8.3} = \frac{1.66 \times 8}{(0.2 \times 8) \times (0.1 \times 83)} = \frac{1.66}{0.2 \times 8.3} = \frac{1.66}{1.66} = 1. \] Simplify the exponential part: \[ \frac{10^{-22}}{10^{-21}} = 10^{-1} \text{ T}. \] \[ B = 1 \times 10^{-1} \text{ T} = 0.1 \text{ T}. \]

Step 5: Convert the result to mT.
Since $1 \text{ T} = 1000 \text{ mT}$: \[ B = 0.1 \times 1000 \text{ mT} = 100 \text{ mT}. \] \[ \boxed{B = 100 \text{ mT}}. \]