Question

A proton moving with velocity \( V \) in a non-uniform magnetic field traces a path as shown in the figure. The path followed by the proton is always in the plane of the paper. What is the direction of the magnetic field in the region near points P, Q, and R? What can you say about relative magnitude of magnetic fields at these points?

Hint:

Use the right-hand rule to determine the direction of the magnetic field based on the direction of motion of a positively charged particle.

Answer 1

Analysis of Proton's Path in a Non-uniform Magnetic Field 

Given:

• A proton moves with velocity \( \vec{V} \) in a plane (paper plane)
• The trajectory is curved, suggesting a non-uniform magnetic field \( \vec{B} \)
• We are to determine the direction and relative magnitude of \( \vec{B} \) at points P, Q, and R.

Key Concepts:

• The magnetic force on a moving charge is given by: \[ \vec{F} = q (\vec{V} \times \vec{B}) \]
• For circular/curved motion: \[ \text{Centripetal force} = \frac{mv^2}{r} = qvB \Rightarrow r = \frac{mv}{qB} \] Thus, a smaller radius indicates a stronger magnetic field.
• Direction of magnetic field can be determined using the right-hand rule: - Fingers in direction of \( \vec{V} \) - Curl toward \( \vec{F} \) (curvature) - Thumb gives \( \vec{B} \)

Answer:

➡ Direction of Magnetic Field:

Since the path of the proton (positive charge) curves to the left, the magnetic force is directed toward the center of curvature. Using the right-hand rule with velocity tangents, we find:

• At all points P, Q, R, the magnetic field is perpendicular to the plane and directed into the page (denoted by a cross “×”).

➡ Relative Magnitude of Magnetic Field:

Since: \[ r = \frac{mv}{qB} \Rightarrow B \propto \frac{1}{r} \] The smaller the radius of curvature, the stronger the magnetic field.

Looking at the image:

• At point Q: radius is smallest ⇒ \( B_Q \) is the strongest.
• At point R: intermediate radius ⇒ \( B_R \) is moderate.
• At point P: largest radius ⇒ \( B_P \) is the weakest.

 

✔ Final Answer:

• Direction of \( \vec{B} \) at P, Q, R: Into the page
• Relative magnitudes: \( B_Q > B_R > B_P \)

Answer 2

Given / idea 

A proton (positive charge) moves with velocity $\vec v$ in a non-uniform magnetic field and its path (always in the plane of the paper) is curved. Magnetic force is the centripetal force that bends the proton’s path: $$\vec F = q(\vec v\times\vec B).$$

Direction of $\vec B$

1. Since the particle’s path lies in the plane of the paper, $\vec B$ must be perpendicular to that plane (either into or out of the page).
2. Use the right-hand rule for $\vec v\times\vec B$ with $q>0$ (proton). If $\vec B$ is into the page and $\vec v$ is tangential, $\vec v\times\vec B$ points toward the centre of curvature — producing the observed bending. (If $\vec B$ were out of the page the force would be in the opposite sense and the curvature would reverse.)
3. Because the observed curvature direction at P, Q and R is consistent with the force produced by $\vec B$ into the page, the magnetic field near all three points is into the page.

Relative magnitudes of $\vec B$ at P, Q and R

For a charged particle moving perpendicular to a magnetic field the radius of curvature $r$ is $$r=\dfrac{mv}{qB}\quad\Rightarrow\quad B=\dfrac{mv}{q\,r}.$$ With the proton’s mass $m$, charge $q$ and speed $v$ (given) essentially the same at those nearby points, the magnetic field strength is inversely proportional to the local radius of curvature: $B\propto 1/r$.

From the figure the curvature radii satisfy (smallest curvature radius at Q, larger at R, largest at P) — therefore $$B_Q > B_R > B_P.$$

Final answer

Direction: $\vec B$ is into the page at P, Q and R.
Relative magnitudes: $\displaystyle B_Q > B_R > B_P.$