The mag. field (B) at the centre of a circular coil of radius "a", through which a current I flows is (a) \(\mathrm{B} \propto \mathrm{c} \mathrm{a}\) (b) \(\mathrm{B} \propto(1 / \mathrm{I})\) (c) \(\mathrm{B} \propto \mathrm{I}\) (d) \(B \propto I^{2}\)

Short Answer

Expert verified
The magnetic field B at the center of the circular coil is given by the formula \(B = \frac{\mu_0I}{2a}\), which shows that B is directly proportional to the current I. Therefore, the correct answer is (c) \(B \propto I\).

Step by step solution

01

Write down the Biot-Savart Law for a current element

The Biot-Savart Law gives the magnetic field produced by a small current element, and is given by: \[ d\vec{B} = \frac{\mu_0}{4\pi} \cdot \frac{I \cdot d\vec{l} \times \vec{r}}{r^3} \] where \(d\vec{B}\) is the magnetic field produced by the current element, \(\mu_0\) is the permeability of free space, I is the current flowing through the current element, \(d\vec{l}\) is the length of the current element, \(\vec{r}\) is the position vector from the current element to the point where we want to find the magnetic field, and r is the magnitude of \(\vec{r}\).
02

Integrate the Biot-Savart Law for the circular coil

The given coil is a closed circular loop, so we need to integrate the Biot-Savart Law along the entire loop to find the total magnetic field at the center. Let's consider the center of the loop to be at the origin (0, 0, 0). The position vector \(\vec{r}\) will have the same magnitude "a" for all points on the loop, and the angle between \(d\vec{l}\) and \(\vec{r}\) at each point will be 90 degrees. So, the cross product simplifies to: \[ d\vec{l} \times \vec{r} = |d\vec{l}| |\vec{r}| \sin{90^{\circ}} = adl \] Now we need to integrate around the entire loop: \[ B = \oint \frac{\mu_0}{4\pi} \cdot \frac{I \cdot adl}{a^3} = \frac{\mu_0}{4\pi} \cdot \frac{I \cdot a}{a^3} \oint dl \] Since the integral of the length element \(dl\) around the whole loop is simply the circumference of the loop (2πa), we have: \[ B = \frac{\mu_0}{4\pi} \cdot \frac{I \cdot a}{a^3} \cdot 2\pi a = \frac{\mu_0I}{2a} \]
03

Derive the correct relation from the result

Now we have found the magnetic field B at the center of the loop: \[ B = \frac{\mu_0I}{2a} \] From this result, we can find the correct proportionality relation among the given options. (a) B ∝ ca: This option implies that B is proportional to the radius a, but here B is inversely proportional to a. (b) B ∝ (1/I): This option implies that B is inversely proportional to I, but here B is directly proportional to I. (c) B ∝ I: This matches our result since B is directly proportional to I. (d) B ∝ I²: This option implies that B is proportional to the square of I, but our result shows it's directly proportional to I, not its square. Therefore, the correct answer is (c) B ∝ I.

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Most popular questions from this chapter

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

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A long straight wire carrying current of \(30 \mathrm{Amp}\) is placed in an external uniform mag. field of induction \(4 \times 10^{-4}\) tesla. The mag. field is acting parallel to the direction of current. The magnitude of the resultant magnetic induction in tesla at a point \(2 \mathrm{~cm}\) away from the wire is tesla. (a) \(10^{-4}\) (c) \(5 \times 10^{-4}\) (b) \(3 \times 10^{-4}\) (d) \(6 \times 10^{-4}\)

Two iclentical short bar magnets, each having magnetic moment \(\mathrm{M}\) are placed a distance of \(2 \mathrm{~d}\) apart with axes perpendicular to each other in a horizontal plane. The magnetic induction at a point midway between them is. (a) $\sqrt{2}\left(\mu_{0} / 4 \pi\right)\left(\mathrm{M} / \mathrm{d}^{3}\right)$ (b) $\sqrt{3}\left(\mu_{0} / 4 \pi\right)\left(\mathrm{M} / \mathrm{d}^{3}\right)$ (c) $\sqrt{4}\left(\mu_{0} / 4 \pi\right)\left(\mathrm{M} / \mathrm{d}^{3}\right)$ (d) $\sqrt{5}\left(\mu_{0} / 4 \pi\right)\left(\mathrm{M} / \mathrm{d}^{3}\right)$

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