Animated Solution for Physics - Magnetic Effects of Current: The figure shows a region of length $l$ with a uniform magnetic field of $0.3\text{ T}$ in it and a proton entering the region with velocity $4\times 10^5{\text{ ms}}^{-1}$ making an angle $60^{\circ }$ with the field. If the proton completes 10 revolutions by the time it cross the region shown, $l$ is close to (Take, mass of proton $=1.67\times 10^{-27}\text{ kg}$, charge of the proton $=1.6\times 10^{-19}\text{ C}$)

Imagine a proton hurtling through space, suddenly entering a region filled with a uniform magnetic field. But it doesn't enter head-on, nor does it enter perfectly sideways. It enters at a daring angle of $60^{\circ }$. This specific angle is the key to unlocking the beautiful, spiraling motion that follows.

Because the velocity vector $\vec{v}$ is at an angle $\theta =60^{\circ }$ to the magnetic field $\vec{B}$, we must break it down into two independent components to understand the proton's fate:

1. The Perpendicular Component ($v_{\perp }$): This component, equal to $v\sin 60^{\circ }$, is entirely responsible for the circular motion. The magnetic force acts as a centripetal force, constantly pulling the proton into a loop. 2. The Parallel Component ($v_{\parallel }$): This component, equal to $v\cos 60^{\circ }$, points in the exact same direction as the magnetic field. Since the magnetic force is given by $\vec{F}=q(\vec{v}\times \vec{B})$, a parallel velocity experiences zero magnetic force. Thus, the proton coasts forward at a constant speed.

Combine a constant circular motion with a constant forward motion, and you get a helix—much like a spiral staircase.

How long does it take for the proton to complete one full loop of this spiral? This is known as the time period, $T$. Remarkably, the time period of a charged particle in a magnetic field depends only on its mass $m$, its charge $q$, and the strength of the magnetic field $B$. It is completely independent of how fast it is going!

The problem states that the proton completes exactly $n=10$ revolutions before it exits the region. Therefore, the total time $t$ it spends inside the magnetic field is simply ten times the time period:

While the proton is busy spinning around 10 times, it is also steadily moving forward along the length of the region, $l$. The distance it covers in this forward direction is dictated solely by its parallel velocity component, $v_{\parallel }$.

Substituting our expressions for $v_{\parallel }$ and $t$, we get our master equation:

Now, we carefully substitute the given values. Don't let the scientific notation intimidate you; handle the numbers and the powers of 10 separately.

Knowing that $\cos 60^{\circ }=0.5$, the forward velocity is $2\times 10^5{\text{ ms}}^{-1}$.

Rounding to two decimal places, we find that the length of the region $l$ is approximately $0.44\text{ m}$. This matches perfectly with option (c). By breaking down a complex 3D motion into two simple 1D and 2D motions, we've elegantly solved the mystery of the proton's helical path!