Problem 75 - Master Resource Book in JEE Main Physics

Question

A straight conductor of length \(l\) carrying a current \(i\), is bent in the form of a semi-circle. The magnetic field in tesla at the centre of the semi-circle is [Kerala CET 2005] (a) \(\frac{\pi^{2} i}{l} \times 10^{-7}\) (b) \(\frac{i \pi}{l} \times 10^{-7}\) (c) \(\frac{\pi i}{I^{2}} \times 10^{-7}\) (d) \(\frac{\pi^{2}}{l} \times 10^{-7}\) (e) None of the above

Step-by-Step Solution

Answer

The magnetic field at the center is \( \frac{\pi^2 i}{l} \times 10^{-7} \text{ T} \), corresponding to option (a).

Step 1: Understanding the Problem

We are given a straight conductor of length \( l \) carrying a current \( i \). This conductor is bent into the shape of a semi-circle. We need to find the magnetic field at the center of the semi-circle formed by the conductor.

Step 2: Recalling the Magnetic Field Formula

The magnetic field at the center of a circular loop due to current \( i \) is given by the formula \( B = \frac{\mu_0 i}{2R} \), where \( R \) is the radius of the circle, and \( \mu_0 \) is the permeability of free space: \( 4 \pi \times 10^{-7} \text{ Tm/A} \). In a semi-circle, the formula becomes \( B = \frac{\mu_0 i}{4R} \).

Step 3: Finding the Radius

Since the wire is bent into a semi-circle, the length of this semi-circle is \( l = \pi R \). Hence, the radius \( R \) of the semi-circle is \( R = \frac{l}{\pi} \).

Step 4: Substitute the Radius into Magnetic Field Formula

Substitute \( R = \frac{l}{\pi} \) into the formula for the magnetic field at the center of a semi-circle: \( B = \frac{\mu_0 i}{4(\frac{l}{\pi})} = \frac{\mu_0 i \pi}{4l} \).

Step 5: Substitute Values to Get the Result

Substitute \( \mu_0 = 4 \pi \times 10^{-7} \) into the expression: \( B = \frac{4 \pi \times 10^{-7} i \pi}{4l} = \frac{\pi^2 i}{l} \times 10^{-7} \).

Step 6: Conclusion

From the derived formula, the correct answer for the magnetic field at the center of the semi-circle is: \( \frac{\pi^2 i}{l} \times 10^{-7} \text{ T} \). This corresponds with option (a) in the given choices.

Key Concepts

Current-Carrying Conductor

A current-carrying conductor is any wire or object through which electric current flows. The current generates a magnetic field around the conductor. This is a fundamental concept in electromagnetism.

• Electric currents create magnetic effects, such as magnetic fields.
• Understanding this is crucial for working with electromagnets, motors, and transformers.

This principle is seen in actions like a simple straight conductor, which produces a circular magnetic field around it. The magnetic field's direction follows the Right Hand Rule: if the thumb of your right hand points in the current's direction, the fingers wrap around in the direction of the magnetic field lines.
By bending the conductor into different shapes, like a semi-circle, the magnetic field pattern also changes, affecting how strong the field is at certain points.

Semi-Circle Wire

When a straight wire is bent into the shape of a semi-circle, the geometry of its magnetic field changes. It's important because it helps us understand how varying wire shapes affect magnetic fields.

• A semi-circle is half of a circle, impacting how the magnetic field is calculated.
• The geometry affects the field's spread and intensity.

To calculate the magnetic field at the center of such a semi-circle, we start by finding its radius. Given that the wire's length replaces the circle's circumference, and in a semi-circle: \[ l = rac{ ext{circumference of full circle}}{2} = rac{2 ext{π}R}{2} = ext{π}R \] Solving for the radius of the semi-circle yields:\[ R = rac{l}{ ext{π}} \]Now, with this radius, we can use it in magnetic field calculations specific to a semi-circle.

Magnetic Field at Center

The magnetic field at the center of a semi-circle is vital in understanding how bending a wire impacts magnetic forces. It is a specific case of applying magnetic field concepts to practical shapes. When dealing with a semi-circle:

• The center of the semi-circle acts as a special point of interest for calculating the field strength.
• Only half of the circular loop contributes to the field.

For a semi-circle, the magnetic field formula becomes:\[ B = \frac{ \mu_0 i}{4R} \]After substituting the radius \( R \) as \( \frac{l}{\text{π}} \), we can give the expression specific to our semi-circle wire:\[ B = \frac{ \mu_0 i \text{π}}{4l} \]This result shows us how rearranging the wire into a semi-circle modifies the magnetic field strength at its center.

Permeability of Free Space

The permeability of free space, denoted by \( \mu_0 \), is a constant in electromagnetism, crucial for calculating magnetic fields. It's measured in tesla meters per ampere (Tm/A).

• \( \mu_0 \) provides the goodness factor for magnetic field calculations.
• It is a fundamental constant used to relate electric currents with magnetic fields.

Numerically, \( \mu_0 = 4 \pi \times 10^{-7} \) Tm/A. In the semi-circle problem, we used this constant to predict the magnetic field at the center of the circle:\[ B = \frac{4 \pi \times 10^{-7} i \text{π}}{4l} = \frac{\text{π}^2 i}{l} \times 10^{-7} \text{ T} \]Thus, \( \mu_0 \) is essential for turning the theoretical expressions into quantitative predictions of magnetic field strengths.