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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

A magnet of magnetic moment \(50 \uparrow \mathrm{A} \mathrm{m}^{2}\) is placed along the \(\mathrm{X}\) -axis in a mag. field $\mathrm{B}^{-}=(0.5 \uparrow+3.0 \mathrm{~J} \wedge$ ) Tesla. The torque acting on the magnet is N.m. (c) \(75 \mathrm{k} \wedge\) (d) \(25 \sqrt{5} \mathrm{k} \wedge\) (a) \(175 \mathrm{k}\) (b) \(150 \mathrm{k}\)

Two particles \(\mathrm{X}\) and \(\mathrm{Y}\) having equal charges, after being accelerated through the same potential difference, enter a region of uniform mag. field and describe circular path of radius \(\mathrm{r}_{1}\) and \(\mathrm{r}_{2}\) respectively. The ratio of mass of \(\mathrm{X}\) to that of \(\mathrm{Y}\) is (a) \(\sqrt{\left(r_{1} / \mathrm{r}_{2}\right)}\) (b) \(\left(\mathrm{r}_{2} / \mathrm{r}_{1}\right)\) (c) \(\left(\mathrm{r}_{1} / \mathrm{r}_{2}\right)^{2}\) (b) \(\left(\mathrm{r}_{1} / \mathrm{r}_{2}\right)\)

An electron having mass \(9 \times 10^{-31} \mathrm{~kg}\), charge $1.6 \times 10^{-19} \mathrm{C}\( and moving with a velocity of \)10^{6} \mathrm{~m} / \mathrm{s}$ enters a region where mag. field exists. If it describes a circle of radius \(0.10 \mathrm{~m}\), the intensity of magnetic field must be Tesla (a) \(1.8 \times 10^{-4}\) (b) \(5.6 \times \overline{10^{-5}}\) (c) \(14.4 \times 10^{-5}\) (d) \(1.3 \times 10^{-6}\)

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 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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