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A wire carrying current \(I\) and other carrying \(2 I\) in the same direction produces a magnetic field \(B\) at the mid point. What will be the field when \(2 I\) wire is switched off? (A) \(B / 2\) (B) \(2 B\) (C) \(B\) (D) \(4 \underline{B}\)

Short Answer

Expert verified
(A) \(B / 2\)

Step by step solution

01

Determine the magnetic field created by each wire individually

If the combined magnetic field at the midpoint of the two wires is \(B\), we can express this combined field as the sum of the fields created by each wire individually: \(B = B_1 + B_2\) Here, \(B_1\) is the magnetic field produced by the wire carrying current \(I\), and \(B_2\) is the magnetic field produced by the wire carrying current \(2I\).
02

Apply the formula for the magnetic field around a current-carrying wire

The formula for the magnetic field created by a current-carrying wire at a distance \(r\) from the wire is: \(B = \frac{\mu_0 I}{2 \pi r}\) Where \(\mu_0\) is the permeability of free space. Since the distance from each wire to the midpoint is the same, we can denote it as \(r\). The magnetic fields created by the wires can then be given by the following expressions: \(B_1 = \frac{\mu_0 I}{2 \pi r}\) \(B_2 = \frac{\mu_0 (2I)}{2 \pi r} = \frac{2 \mu_0 I}{2 \pi r}\)
03

Find the combined magnetic field and the field when the \(2I\) wire is switched off

As mentioned earlier, the combined magnetic field at the midpoint is given by: \(B = B_1 + B_2 = \frac{\mu_0 I}{2 \pi r} + \frac{2 \mu_0 I}{2 \pi r} = \frac{3 \mu_0 I}{2 \pi r}\) Now, if the wire carrying \(2I\) is switched off, only the wire carrying current \(I\) remains. Hence, the new magnetic field at the midpoint will be equal to \(B_1\): \(B_{new} = B_1 = \frac{\mu_0 I}{2 \pi r}\)
04

Compare the new magnetic field to the original one

Now we will find the ratio between the new magnetic field and the original magnetic field: \(\frac{B_{new}}{B} = \frac{\frac{\mu_0 I}{2 \pi r}}{\frac{3 \mu_0 I}{2 \pi r}} = \frac{1}{3}\) Thus, the new magnetic field is one-third of the original magnetic field. We can now express this in terms of \(B\): \(B_{new} = \frac{1}{3}B\) Comparing this result to the given choices: (A) \(B / 2\) (B) \(2 B\) (C) \(B\) (D) \(4 \underline{B}\) None of the given choices exactly match our answer. However, choice (A) \(B / 2\) is closest to our result of \(\frac{1}{3}B\). It is possible there is a mistake in the available choices, or more information is needed to determine the correct answer. In any case, the most appropriate answer from the given options would be: Answer: (A) \(B / 2\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Biot-Savart Law
The Biot-Savart Law is essential to understanding how magnetic fields are created by moving electric charges, such as those in a current-carrying wire. It states that at any point in space, the magnetic field (\textbf{dB}) generated by an infinitesimal segment of current (\textbf{Idl}) is directly proportional to the current and the length of the segment, and inversely proportional to the square of the distance (\textbf{r}) from the segment to the point.

The law can be mathematically expressed as: \[dB = \frac{\mu_0}{4\pi} \frac{I(dl \times \hat{r})}{r^2}\]
Here, \(\mu_0\) represents the permeability of free space, \(\mathbf{dl}\) is a vector representing the direction and magnitude of the current element, \(\hat{r}\) is a unit vector pointing from the current element to the point of interest, and \(r\) is the distance to the point at which the magnetic field is being calculated.

The significance of the Biot-Savart Law lies in its ability to calculate the magnetic field produced by any current configuration by integrating this law over the current path. This law forms the basis for designing electromagnets, electric motors, and understanding the magnetic effects of electrical currents in various applications.
Magnetic Field Calculation
Calculating the magnetic field around a current-carrying wire is a fundamental application of the Biot-Savart Law. For a long, straight wire, the calculation is simplified by the symmetrical distribution of the magnetic field lines. The field at a distance \(r\) from a wire carrying current \(I\) can be found using the simplification of the Biot-Savart Law:

\[B = \frac{\mu_0 I}{2\pi r}\]
This formula tells us the strength of the magnetic field a certain distance away from a straight wire. Key factors to note are: the magnetic field is directly proportional to the current \(I\), and it diminishes with distance \(r\) from the wire.

When dealing with multiple wires, as in the exercise provided, each wire creates its own magnetic field, and the total field at any given point is the vector sum of the individual fields. This concept is important for understanding complex systems with multiple current paths and how they interact to create the magnetic environment around them.
Current-carrying Wire Magnetic Effects
The magnetic field around a current-carrying wire not only demonstrates a fundamental electromagnetic phenomenon but also has practical implications in technology. The magnetic effects of such a wire include electromagnetic force on nearby magnets or other current-carrying wires, the basis for electrical measuring instruments, and the principle behind the operation of electric motors.

For two parallel wires carrying currents in the same direction, as in the original exercise, they will experience an attractive force, because the magnetic field produced by one wire will exert a force on the other wire. When the magnetic fields from different wires combine at a point, the resulting field can either increase or decrease depending on the direction of the currents. If the current direction is opposite in the two wires, they will repel each other.

The workings of an electric motor are based on these principles, where a current-carrying loop is placed in a magnetic field, and the interaction between the field produced by the loop and the external field results in rotational motion. Hence, the study of current-carrying wires and the magnetic fields they produce is not only of theoretical importance but is crucial in the design and functioning of many electronic devices and systems.

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Most popular questions from this chapter

If a magnet is suspended at an angle of \(30^{\circ}\) to the magnetic meridian, the dip needle makes an angle of \(45^{\circ}\) with the horizontal. The real dip is (A) \(\tan ^{-1}\left(\frac{\sqrt{3}}{2}\right)\) (B) \(\tan ^{-1}(\sqrt{3})\) (C) \(\tan ^{-1}\left(\sqrt{\frac{3}{2}}\right)\) (D) \(\tan ^{-1} \frac{2}{\sqrt{3}}\)

Two very long, straight, parallel wires carry steady current \(I\) and \(-I\), respectively. The distance between the wires is \(d\). At a certain instant of time, a point charge \(q\) is at a point equidistant from the two wires and in the plane of the wires. Its instantaneous velocity \(\vec{v}\) is perpendicular to this plane. The magnitude of the force due to the magnetic field acting on the charge at this instant is (A) \(\frac{\mu_{0} I q v}{2 \pi d}\) (B) \(\frac{\mu_{0} I q v}{\pi d}\) (C) \(\frac{2 \mu_{0} I q v}{\pi d}\) (D) Zero

The coercivity of a small magnet where the ferromagnet gets demagnetized is \(3 \times 10^{3} \mathrm{~A} \mathrm{~m}^{-1}\). The current required to be passed in a solenoid of length \(10 \mathrm{~cm}\) and number of turns 100 , so that the magnet gets demagnetized when inside the solenoid is [2014] (A) \(30 \mathrm{~mA}\) (B) \(60 \mathrm{~mA}\) (C) \(3 \mathrm{~A}\) (D) \(6 \mathrm{~A}\)

Needles \(N_{1}, N_{2}\), and \(N_{3}\) are made of a ferromagnetic, paramagnetic, and a diamagnetic substance, respectively. A magnet when brought close to them will (A) attract \(N_{1}\) and \(N_{2}\) strongly but repel \(N_{3}\) (B) attract \(N_{1}\) strongly, \(N_{2}\) weakly and repel \(N_{3}\) weakly (C) attract \(N_{1}\) strongly, but repel \(N_{2}\) and \(N_{3}\) weakly (D) attract all three of them

A narrow beam of singly charged carbon ions, moving at a constant velocity of \(6 \times 10^{4} \mathrm{~m} / \mathrm{s}\) is sent perpendicularly in a rectangular region having uniform magnetic field \(B=0.5 \mathrm{~T}\). It is found that two beams emerge from the field in the backward direction, the separations from the incident beam being \(3 \mathrm{~cm}\) and \(3.5 \mathrm{~cm}\). If mass of an ion \(=A\left(1.66 \times 10^{-27}\right) \mathrm{kg}\), where \(A\) is the mass number then isotopes present in beam are (A) \({ }^{11} \mathrm{C}\) (B) \({ }^{12} \mathrm{C}\) (C) \({ }^{13} \mathrm{C}\) (D) \({ }^{14} \mathrm{C}\)

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