Problem 5 - Master Resource Book in JEE Main Physics

Question

The wing span of an aeroplane is \(36 \mathrm{~m}\). If the plane is flying at \(400 \mathrm{kmh}^{-1}\), the emf induced between the wings tips is (Assume \(V=4 \times 10^{-5} \mathrm{~T}\) ) (a) \(16 \mathrm{~V}\) (b) \(1.6 \mathrm{~V}\) (c) \(0.16 \mathrm{~V}\) (d) \(0.016 \mathrm{~V}\)

Step-by-Step Solution

Answer

The correct answer is (c) \(0.16 \mathrm{~V}\).

Step 1: Understanding the Given Values

We are given: Wing span \( l = 36 \mathrm{~m} \), Velocity of the plane \( v = 400 \mathrm{~kmh}^{-1} \), and Magnetic field \( B = 4 \times 10^{-5} \mathrm{~T} \). The formula for induced emf is \( E = B \cdot l \cdot v \).

Step 2: Convert Units

First, convert the velocity from \( \mathrm{kmh}^{-1} \) to \( \mathrm{ms}^{-1} \). We know \( 1 \mathrm{~kmh}^{-1} = \frac{1}{3.6} \mathrm{~ms}^{-1} \). So, \( v = 400 \times \frac{1}{3.6} \approx 111.11 \mathrm{~ms}^{-1} \).

Step 3: Calculate the Induced EMF

Use the formula \( E = B \cdot l \cdot v \). Substitute the values we have: \( E = 4 \times 10^{-5} \mathrm{~T} \times 36 \mathrm{~m} \times 111.11 \mathrm{~ms}^{-1} \).

Step 4: Solve the Equation

Calculate: \( E = 4 \times 10^{-5} \times 36 \times 111.11 \). This simplifies to \( E = 0.16 \mathrm{~V} \).

Step 5: Verify Against Options

The calculated induced emf \( 0.16 \mathrm{~V} \) matches option (c).

Key Concepts

Induced EMF

Electromagnetic induction is a fascinating phenomenon where a change in magnetic field induces an electromotive force (EMF) in a conductor. Faraday's Law of Induction underpins this concept, explaining how a changing magnetic environment creates an electrical current. This concept is visually grasped through a simple formula:\[ E = B \cdot l \cdot v \]where:

• \( E \) is the induced EMF, which is measured in volts (V).
• \( B \) is the magnetic flux density, measured in teslas (T).
• \( l \) signifies the length of the conductor, such as the wingspan of an airplane, in meters (m).
• \( v \) represents the velocity of the conductor through the magnetic field, measured in meters per second (m/s).

In our exercise, as the aeroplane moves at a certain speed through the Earth's magnetic field, a potential difference is induced between the wing tips. Understanding this principle allows us to predict how various factors, such as speed and magnetic field strength, impact the resulting EMF.

Magnetic Field

A magnetic field can be viewed as an invisible force altering space around magnetic materials or electric currents. It's a vector field, meaning it has both direction and magnitude, and it profoundly influences how charged particles move. The strength of a magnetic field is often described by magnetic flux density, denoted as \( B \), measured in teslas (T). In the context of electromagnetic induction, the magnetic field's interaction with a conductor manifests as an induced EMF. The airplane in our exercise flies through a weak magnetic field of \( 4 \times 10^{-5} \, \mathrm{T}\). This implies that even a subtle ambient magnetic field can generate an EMF when a conductor moves at noticeable velocity relative to it. This showcases the importance of magnetic fields in various technological applications, from designing motors to ensuring navigation systems function correctly in aviation.

Unit Conversion in Physics

Unit conversion is an essential skill within physics, allowing us to understand and solve problems using consistent units. In the given exercise, converting the airplane's speed from kilometers per hour (km/h) to meters per second (m/s) is crucial for calculating the induced EMF correctly.To achieve this conversion, one must remember that:1 kilometer per hour = \( \frac{1}{3.6} \) meters per second.Thus, for a given velocity of \( 400 \, \mathrm{kmh}^{-1} \) in the exercise:

• We convert it to \( 111.11 \, \mathrm{ms}^{-1} \) using the formula: \( 400 \times \frac{1}{3.6} \approx 111.11 \).

Such conversions ensure all units align properly, facilitating accurate computations. Consistent unit application helps unify the problem-solving process across physics, making complex equations and concepts easier to handle and understand.