For the mag. field to be maximum due to a small element of current carrying conductor at a point, the angle between the element and the line joining the element to the given point must be (a) \(0^{\circ}\) (b) \(90^{\circ}\) (c) \(180^{\circ}\) (d) \(45^{\circ}\)

Short Answer

Expert verified
For the magnetic field to be maximum due to a small element of current-carrying conductor at a point, the angle between the element and the line joining the element to the given point must be \(90^{\circ}\), according to the Biot-Savart law. Therefore, the correct answer is (b) \(90^{\circ}\).

Step by step solution

01

Understand Biot-Savart law

The Biot-Savart law helps us find the magnetic field (dB) produced by a small current element (Idl). It states that the magnitude of the magnetic field produced is directly proportional to the current (I), the length of the small current element (dl), the sine of the angle (θ) between the line joining the element and the given point, and inversely proportional to the square of the distance (r) from the element to the given point. The formula for Biot-Savart law is given by: \( dB = \frac{\mu_0}{4\pi} \frac{I(dl \times r \sin(\theta))}{r^2} \) where µ₀ is the permeability of free space.
02

Analyzing the formula for maximum magnetic field

In order to maximize dB, we need to find the maximum value of \(\sin(\theta)\). Since the sine function has a maximum value of 1, this occurs when the angle θ is equal to \(90^{\circ}\). The formula then simplifies to: \( dB = \frac{\mu_0}{4\pi} \frac{I(dl \times r)}{r^2} \)
03

Identifying the correct angle

As we see from the analysis in Step 2, the maximum magnetic field occurs when \(\theta = 90^{\circ}\). Therefore, the answer is (b) \(90^{\circ}\).

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Most popular questions from this chapter

Needles \(\mathrm{N}_{1}, \mathrm{~N}_{2}\) and \(\mathrm{N}_{3}\) are made of a ferromagnetic, a paramagnetic and a diamagnetic substance respectively. A magnet when brought close to them will (a) Attract \(\mathrm{N}_{1}\) strongly, \(\mathrm{N}_{2}\) weakly and repel \(\mathrm{N}_{3}\) weakly (b) Attract \(\mathrm{N}_{1}\) strongly, but repel \(\mathrm{N}_{2}\) and \(\mathrm{N}_{3}\) weakly (c) Attract all three of them (d) Attract \(\mathrm{N}_{1}\) and \(\mathrm{N}_{2}\) strongly but repel \(\mathrm{N}_{2}\)

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) $\sqrt{5}\left(\mu_{0} / 4 \pi\right)\left(\mathrm{M} / \mathrm{d}^{3}\right)$

A conducting rod of length \(\ell\) [cross-section is shown] and mass \(\mathrm{m}\) is moving down on a smooth inclined plane of inclination \(\theta\) with constant speed v. A vertically upward mag. field \(\mathrm{B}^{-}\) exists in upward direction. The magnitude of mag. field \(B^{-}\) is(a) $[(\mathrm{mg} \sin \theta) /(\mathrm{I} \ell)]$ (b) \([(\mathrm{mg} \cos \theta) /(\mathrm{I} \ell)]\) (c) \([(\mathrm{mg} \tan \theta) /(\mathrm{I} \ell)]\) (d) \([(\mathrm{mg}) /(\mathrm{I} \ell \sin \theta)]\)

The strength of the magnetic field at a point \(\mathrm{y}\) near a long straight current carrying wire is \(\mathrm{B}\). The field at a distance \(\mathrm{y} / 2\) will be (a) B/2 (b) B \(/ 4\) (c) \(2 \mathrm{~B}\) (d) \(4 \mathrm{~B}\)

Two thin long parallel wires separated by a distance \(\mathrm{y}\) are carrying a current I Amp each. The magnitude of the force per unit length exerted by one wire on other is (a) \(\left[\left(\mu_{0} I^{2}\right) / y^{2}\right]\) (b) \(\left[\left(\mu_{o} I^{2}\right) /(2 \pi \mathrm{y})\right]\) (c) \(\left[\left(\mu_{0}\right) /(2 \pi)\right](1 / y)\) (d) $\left[\left(\mu_{0}\right) /(2 \pi)\right]\left(1 / \mathrm{y}^{2}\right)$

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