Proton with kinetic energy of 1 MeV moves from south to north. It gets an acceleration of 10^12 m/s^

Q: Why is the angle $\theta$ taken as $90^\circ$?

The proton moves from South to North, while the magnetic field points from West to East. These two directions are perpendicular to each other, making the angle exactly $90^\circ$.

Q: Why do we need to convert MeV to Joules?

To use standard physics formulas like $F=ma$ and $K=\frac{1}{2}mv^2$ correctly, all units must be in the standard SI system. Joules is the SI unit for energy, while MeV is not.

Q: What does the cross symbol ($\otimes$) mean in the diagram?

In physics, a circle with a cross ($\otimes$) represents a vector pointing directly into the page (like the tail feathers of a dart). A circle with a dot ($\odot$) represents a vector pointing out of the page.

Q: Would the answer change if it was an electron?

The magnitude of the force would be the same if the velocity was the same, but the direction of the force would be opposite (out of the page) because an electron has a negative charge. Also, an electron's mass is much smaller, so its velocity for the same kinetic energy would be much higher!

Q: How did $\frac{1}{\sqrt{2}}$ become $0.707$?

The value of $\sqrt{2}$ is approximately $1.414$. If you divide $1$ by $1.414$, you get approximately $0.707$. It's a very common fraction in physics and trigonometry!

Question

Proton with kinetic energy of $1 	ext{ MeV}$ moves from south to north. It gets an acceleration of $10^{12} 	ext{ m/s}^2$ by an applied magnetic field (west to east). The value of magnetic field (rest mass of proton is $1.6 	imes 10^{-27} 	ext{ kg}$)

Options:

(A) $71 	ext{ mT}$

(B) $0.071 	ext{ mT}$

(C) $0.71 	ext{ mT}$

(D) $7.1 	ext{ mT}$

Sigma Percentile · Accepted Answer

Imagine a proton, a tiny positively charged particle, cruising steadily from South to North. It's moving with a kinetic energy of $1 	ext{ MeV}$. Suddenly, it enters a region where an invisible magnetic field is pointing from West to East. What happens next is a beautiful dance of physics!

### The Setup: A Proton's Journey
When a charged particle moves through a magnetic field, it doesn't just pass through unaffected. It experiences a force. According to the right-hand rule, if you point your fingers in the direction of the velocity (North) and curl them towards the magnetic field (East), your thumb points downwards. This means the magnetic force pushes the proton directly into the plane of its motion!

### The Invisible Hand: Magnetic Force
The magnitude of this magnetic force is given by the elegant equation:
$$F_m = qvB \sin(	heta)$$

Since the proton is moving perpendicular to the magnetic field, $	heta = 90^\circ$, and $\sin(90^\circ) = 1$. So, the force simplifies to $F_m = qvB$.

But wait, the problem tells us that this force gives the proton an acceleration of $10^{12} 	ext{ m/s}^2$. Enter Newton's second law! We know that any net force equals mass times acceleration:
$$F = ma$$

By equating these two forces, we get a powerful relationship:
$$ma = qvB$$

Since we are on a quest to find the magnetic field $B$, let's rearrange this equation:
$$B = \frac{ma}{qv}$$

### Bridging the Gap: From Energy to Velocity
We have the mass $m$, the acceleration $a$, and the charge $q$. But we are missing the velocity $v$! Don't panic. The problem gave us the proton's kinetic energy, $K = 1 	ext{ MeV}$.

First, let's convert this energy into standard SI units (Joules). We know that $1 	ext{ eV} = 1.6 	imes 10^{-19} 	ext{ J}$, so:
$$K = 10^6 	imes 1.6 	imes 10^{-19} 	ext{ J} = 1.6 	imes 10^{-13} 	ext{ J}$$

Now, recall the classic kinetic energy formula:
$$K = \frac{1}{2}mv^2$$

Rearranging for velocity, we get:
$$v = \sqrt{\frac{2K}{m}}$$

Let's plug in the numbers:
$$v = \sqrt{\frac{2 	imes 1.6 	imes 10^{-13}}{1.6 	imes 10^{-27}}}$$

Notice how beautifully the $1.6$ cancels out! We are left with:
$$v = \sqrt{2 	imes 10^{14}} = \sqrt{2} 	imes 10^7 	ext{ m/s}$$

### The Final Calculation: Putting It All Together
Now that we have the velocity, we can finally find the magnetic field. Let's bring back our equation for $B$:
$$B = \frac{ma}{qv}$$

Substitute all our known values:
$$B = \frac{1.6 	imes 10^{-27} 	imes 10^{12}}{1.6 	imes 10^{-19} 	imes \sqrt{2} 	imes 10^7}$$

Once again, the $1.6$ in the numerator and denominator perfectly cancel each other out. Let's combine the powers of 10:
$$B = \frac{10^{-15}}{\sqrt{2} 	imes 10^{-12}} = \frac{1}{\sqrt{2}} 	imes 10^{-3} 	ext{ T}$$

Since $\frac{1}{\sqrt{2}} \approx 0.707$, we can write:
$$B \approx 0.707 	imes 10^{-3} 	ext{ T} = 0.707 	ext{ mT}$$

Rounding to two significant digits, we get our final answer: **$0.71 	ext{ mT}$**.

Physics is all about connecting the dots. By linking kinetic energy to velocity, and magnetic force to Newton's laws, we unveiled the invisible magnetic field steering our proton!

Animated Solution for Physics - Magnetic Effects of Current: Proton with kinetic energy of $1 MeV$ moves from south to north. It gets an acceleration of $1 0^{12} m/s^{2}$ by an applied magnetic field (west to east). The value of magnetic field (rest mass of proton is $1.6 \times 1 0^{- 27} kg$ )

Select Answer:

Visualized Solution

Magnetic Force on a Moving Charge

Force Equations

Equating Forces

Kinetic Energy to Velocity

Calculating Velocity

Substituting Values

Final Answer

The Way Forward

The Sigma Insight: Motion of a Charge in Magnetic Fields

The Setup

The Invisible Hand

Bridging the Gap

The Final Calculation

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