Problem 1885 - JEE Physics

Question

A straight wire of mass \(200 \mathrm{gm}\) and length \(1.5\) meter carries a current of 2 Amp. It is suspended in mid-air by a uniform horizontal magnetic field B. [take \(\left.\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right]\). The \(\mathrm{B}\) is (a) \(\overline{(2 / 3) \text { tes } 1 a}\) (b) \((3 / 2)\) tesla (c) \((20 / 3)\) tesla (d) (3/20) tesla

Step-by-Step Solution

Answer

The magnetic field B required to suspend the wire in mid-air is (a) \(\frac{2}{3} \text{ tesla}\).

Step 1: Convert the mass of the wire to kg

Since the mass of the wire is given in grams, we need to convert it to kg for our calculations. To do so, divide the mass by 1000. \(m = \frac{200}{1000} kg = 0.2 kg\)

Step 2: Calculate the gravitational force acting on the wire

The gravitational force acting on the wire (its weight) can be found using the formula: \(F_g = m \cdot g\) where m is the mass of the wire and g is the acceleration due to gravity (given as 10 m/s²). \(F_g = (0.2 kg)(10 \frac{m}{s^2}) = 2 N\)

Step 3: Write down the formula for the magnetic force on the wire

The magnetic force F_B acting on a current-carrying wire is given by the formula: \(F_B = B \cdot I \cdot L\) where B is the magnetic field, I is the current passing through the wire, and L is the length of the wire.

Step 4: Equate the magnetic force to the gravitational force

Since the magnetic force is suspending the wire in mid-air, it must be equal to the gravitational force acting on the wire: \(F_B = F_g\)

Step 5: Solve for the magnetic field B

Substitute the known values in the equation: \(B \cdot I \cdot L = F_g\) \(B \cdot (2 A) \cdot (1.5 m) = 2 N\) Now, solve for B: \(B = \frac{2 N}{(2 A)(1.5 m)} = \frac{2}{3} T\) So the magnetic field B required to suspend the wire in mid-air is (a) \(\frac{2}{3} \text{ tesla}\).

Key Concepts

Magnetic Field

A magnetic field is a region around a magnetic material or a moving electric charge within which the force of magnetism acts. It can exert a force on other nearby magnets or electric currents. In this exercise, the wire is suspended in mid-air by a magnetic field, which counteracts the gravitational force pulling it down.

The formula to calculate the magnetic force acting on a current-carrying wire is:

• \( F_B = B \cdot I \cdot L \)
• Where \( F_B \) is the magnetic force, \( B \) is the magnetic field, \( I \) is the current, and \( L \) is the length of the wire.

In this context, the magnetic field \( B \) needed to suspend the wire is calculated by equating it to the gravitational force acting on the wire.

Gravitational Force

Gravitational force is the force of attraction between two masses. Here, it refers to the force pulling the wire downwards due to Earth's gravity. The strength of this force can be calculated using the formula:

• \( F_g = m \cdot g \)
• Where \( F_g \) is the gravitational force, \( m \) is the mass of the wire, and \( g \) is the acceleration due to gravity.

In our problem, the gravitational force acting on the wire is \( 2 \text{ N} \), given by substituting \( m = 0.2 \text{ kg} \) and \( g = 10 \text{ m/s}^2 \) in the equation.

Understanding gravitational force in this context is crucial because it must be perfectly countered by the magnetic force to keep the wire suspended.

Current in Wire

Electric current is the flow of electric charge through a conductor, but here it's important due to its interaction with magnetic fields. The wire in our exercise carries a current of \(2 \text{ Amp}\).

The current flowing through the wire plays a significant role when it interacts with a magnetic field, resulting in a magnetic force. By controlling the current, it is possible to manipulate how strong the force acting on the wire will be.

This relationship is seen in the magnetic force equation used (

• \( F_B = B \cdot I \cdot L \)

,

Suspension of Wire

Suspending a wire in mid-air involves balancing both the magnetic force and gravitational force so that they cancel each other out. When these forces are equal, the wire does not experience a net force and remains stationary in the air.

• To achieve suspension: \( F_B = F_g \)
• This means the magnetic force \( B \cdot I \cdot L \) must precisely equal the gravitational force \( m \cdot g \).

This principle is key in applications like magnetic levitation, where objects are made to float by opposing gravitational force with magnetic force. Understanding how to set these equations equal allows for precise control in technological and experimental applications.

In this problem, the calculated magnetic field strength of \(\frac{2}{3} \text{ tesla}\) was exactly what was needed to counteract the wire's weight, successfully achieving suspension.