Question

A proton is accelerated from rest by a potential difference of \(400 .\) V. The proton enters a uniform magnetic field and follows a circular path of radius \(20.0 \mathrm{~cm} .\) Determine the magnitude of the magnetic field.

Short answer

Answer: The magnitude of the magnetic field is approximately \(7.97 \times 10^{-4}\) T.

Step-by-step solution

Identify the known values

We are given: - The potential difference (V) = 400 V - The radius of the circular path (r) = 20 cm = 0.2 m - The charge of a proton (q) = 1.6 × 10^-19 C - The mass of a proton (m) = 1.67 × 10^-27 kg

Calculate the final kinetic energy of the proton

Using the potential difference, we can find the final kinetic energy (K) of the proton as follows: \(K = q \cdot V\) \(K = (1.6 \times 10^{-19}\,\text{C}) \times (400\,\text{V})\) \(K = 6.4 \times 10^{-17}\,\text{J}\)

Calculate the final velocity of the proton

Using the final kinetic energy, we can determine the final velocity (v) of the proton using the following equation: \(\frac{1}{2} mv^2 = K\) Rearranging the equation to solve for v: \(v = \sqrt{\frac{2K}{m}}\) Substituting the values for K and m: \(v = \sqrt{\frac{2 \times 6.4 \times 10^{-17}\,\text{J}}{1.67 \times 10^{-27}\,\text{kg}}}\) \(v \approx 4.38 \times 10^5\,\text{m/s}\)

Determine the magnetic force that causes the circular motion

The magnetic force (F) acting on the proton is equal to the centripetal force required for circular motion. We can represent this as: \(F_{\text{magnetic}} = F_{\text{centripetal}}\) \(qvB = \frac{mv^2}{r}\) Where B is the magnitude of the magnetic field, and r is the radius of the circular path.

Solve for the magnetic field (B)

Now, we can solve for B: \(B = \frac{mv^2}{rqv}\) Substituting the values for m, v, r, and q: \(B = \frac{(1.67 \times 10^{-27}\,\text{kg}) \times (4.38 \times 10^5\,\text{m/s})^2}{(1.6 \times 10^{-19}\,\text{C}) \times (0.2\,\text{m}) \times (4.38 \times 10^5\,\text{m/s})}\) Calculating B: \(B \approx 7.97 \times 10^{-4}\,\text{T}\) Therefore, the magnitude of the magnetic field is approximately \(7.97 \times 10^{-4}\) T.

Key concepts

Proton Acceleration

When a proton is accelerated, it gains energy and speed due to the influence of an electric field. In our problem, the proton starts from rest, which means it initially has no kinetic energy. It is then subjected to a potential difference of 400 volts. The potential difference essentially represents the electric field's work done on the proton, transferring energy to it and causing it to accelerate.

Protons, being positively charged particles, experience a force in the presence of an electric field, moving from regions of higher potential to lower potential. The energy from this movement is converted into kinetic energy. We're able to calculate this gained kinetic energy using the formula:\[ K = q \cdot V \]where:- \( K \) is the kinetic energy,- \( q \) is the charge of the proton (\(1.6 \times 10^{-19} \text{ C}\)), and- \( V \) is the potential difference (400 V).

This formula provides a direct way to determine how much energy a proton receives when accelerated by a specified voltage.

Circular Motion

The circular motion of a proton in a magnetic field is driven by the perpendicular force exerted by the field. After gaining energy from the potential difference, the proton enters a region where a uniform magnetic field is present, causing it to move in a circular path.

When a charged particle, like a proton, moves through a magnetic field, it experiences a magnetic force perpendicular to both the field and its velocity. This force doesn't change the speed of the proton but continually alters its direction, causing circular motion. The radius of this path (given in our problem as 20 cm or 0.2 m) helps us understand how strong the magnetic influence is.
The relationship governing this motion is given by:\[ F_{\text{magnetic}} = F_{\text{centripetal}} \]where the magnetic force provides the necessary centripetal force to maintain the proton's circular path.

Kinetic Energy

Kinetic energy refers to the energy a body possesses due to its motion. For a proton accelerated by a potential difference, this energy is gained as the proton moves from rest to a specific velocity. In the exercise, we used the equation:\[ K = \frac{1}{2} m v^2 \]to link the proton's kinetic energy to its velocity (\( v \)) and mass (\( m \), where \( m = 1.67 \times 10^{-27} \text{ kg} \)). This equation is crucial because it provides a pathway from the energy gained through an electric field to the velocity of the proton.- We first calculate the kinetic energy using the first formula.- We then rearrange the second equation to solve for velocity, \( v \), which gives us insight into how fast the proton is moving after being accelerated by the electric field.

This process illustrates the conversion of electric potential energy into kinetic energy, a fundamental concept in understanding how fields influence charged particles.

Centripetal Force

Centripetal force is essential for maintaining circular motion. It acts as the inward force that keeps a proton, or any object, moving in a curved path rather than in a straight line. In this scenario, the necessary centripetal force that keeps the proton in a circular path is provided by the magnetic field through which the proton moves. The expression linking these concepts is:\[ F_{\text{centripetal}} = \frac{m v^2}{r} \]where:- \( m \) is the mass of the proton,- \( v \) is its velocity, and- \( r \) is the radius of the circular path.

The centripetal force here comes from the magnetic force \( (qvB) \) acting perpendicularly to the proton's velocity, creating the necessary inward acceleration to maintain circular motion. Understanding this relationship not only helps solve the exercise but also deepens knowledge of how charged particles behave in magnetic fields.