Question

A conductor of length \(\ell\) is placed horizontally perpendicular to a horizontal uniform magnetic field B. Suddenly a charge \(Q\) is passed through it and it is found to jump to height h. Choose the correct relation: (Take acceleration due to gravity g) (1) \(h=\frac{B^{2} \ell^{2} Q^{2}}{m^{2} g}\) (2) \(\mathrm{h}=\frac{2 \mathrm{~B}^{2} \ell^{2} \mathrm{Q}^{2}}{\mathrm{~m}^{2} \mathrm{~g}}\) (3) \(\mathrm{h}=\frac{\mathrm{B}^{2} \ell^{2} \mathrm{Q}^{2}}{2 \mathrm{~m}^{2} \mathrm{~g}}\) (4) \(\mathrm{h}=\frac{2 \mathrm{~B}^{2} \ell^{2} \mathrm{Q}^{2}}{\mathrm{mg}}\)

Short answer

The correct relation is (3): \ \h = \ \frac{ B^2 \ \ell^2 \ \Q^2}{2 \ \m^2 \ \g} \

Step-by-step solution

- Understand the problem

Given a conductor of length \( \ell \ \) placed perpendicular to a uniform magnetic field B, a charge \( Q \ \) flows through it, which causes it to jump to height h. The goal is to determine the correct relationship between these variables.

- Use the Lorentz force

The force \( F \ \) on the conductor due to the magnetic field can be given by the Lorentz force: \[ F = B \ \ell \ Q \]

- Apply Newton's second law

This force will cause an acceleration \( a \ \) in the conductor. According to Newton's second law: \[ F = m \ \cdot a \]

- Calculate acceleration

Using the expressions from Step 2 and Step 3: \[ a = \ \frac{F}{m} = \ \frac{B \ \ell \ Q}{ m} \]

- Relate acceleration to height

The height \( h \ \) the conductor jumps to can be related to the acceleration using kinematic equations. From the equation of motion: \[ h = \ \frac{1}{2} \ \frac{{(B \ \ell \ Q)}^2}{m^2 g} \]

Key concepts

Lorentz force

The Lorentz force is the force exerted on a charged particle moving through a magnetic and electric field. In our case, the magnetic field is uniform and perpendicular to the conductor, and the electric charge Q moves through the conductor. The formula for the Lorentz force when only a magnetic field is present is: \( F = B \ \ell \ Q \) \ \( \) Here, F is the magnetic force, B is the magnetic field strength, \ell\ is the length of the conductor, and Q is the electric charge. This relationship is fundamental because it tells us how the motion of charged particles in magnetic fields creates force, driving various physical phenomena.

kinematic equations

Kinematic equations describe the motion of objects under the influence of forces. These equations help us understand how the height h the conductor jumps to relates to its initial velocity, time, and acceleration. Since the conductor initially starts from rest and jumps due to a force, the kinematic equation we'll use is: \( h = \frac{at^2}{2} \) \ But in our problem, we first derive acceleration (a) from Newton's second law. After finding acceleration, we use the kinematic equation for the distance (height h) traveled under constant acceleration.\ Inserting the acceleration we calculated into our kinematic equation provides us with the relationship between h, the magnetic field B, the conductor length \ell, the charge Q, and other constants like mass (m) and gravitational acceleration (g).

Newton's second law

Newton's second law explains how the force applied to an object relates to its mass and acceleration. It states that: \( F = m \ \cdot a \) \ Utilizing this law, we set the magnetic force (F = B\ \ell\ Q) equal to the product of mass (m) and acceleration (a). This helps us write: \( a = \frac{B \ell \ Q}{m} \) \ From here, we use this acceleration value to then derive the height h using kinematic equations, bridging the gap between forces, mass, acceleration, and the resulting motion. Understanding Newton's law is crucial as it forms the backbone of classical mechanics and is pivotal for solving motion problems like this one.