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The magnetic induction at a point \(P\) which is at a distance \(4 \mathrm{~cm}\) from a long current carrying wire is \(10^{-8}\) tesla. The field of induction at a distance \(12 \mathrm{~cm}\) from the same current would be tesla. (a) \(3.33 \times 10^{-9}\) (b) \(1.11 \times 10^{-4}\) (c) \(3 \times 10^{-3}\) (d) \(9 \times 10^{-2}\)

Short Answer

Expert verified
The magnetic induction at a distance of 12 cm from the same current-carrying wire is \(3.33 \times 10^{-9}\) tesla.

Step by step solution

01

Write down the formula for the magnetic field of a long current-carrying wire

The formula for the magnetic field B produced by a long current-carrying wire is given by: \[B = \frac{\mu_0 I}{2\pi r}\] where μ₀ is the permeability of free space, I is the current through the wire, and r is the distance from the wire.
02

Calculate the magnetic field at 3 times the distance

We need to find the magnetic induction when the distance r is tripled. Let B₁ be the magnetic induction at a distance r and B₂ be the magnetic induction at a distance 3r. By substituting 3r in place of r, we get: \[B_2 = \frac{\mu_0 I}{2\pi(3r)}\]
03

Find the ratio of B₂ to B₁

Now, we will find the ratio of B₂ to B₁: \[\frac{B_2}{B_1} = \frac{\frac{\mu_0 I}{2\pi(3r)}}{\frac{\mu_0 I}{2\pi r}}\]
04

Simplify the ratio

Simplify the expression by canceling the common terms in the numerator and denominator: \[\frac{B_2}{B_1} = \frac{r}{3r} = \frac{1}{3}\]
05

Find the new magnetic induction

We know that the magnetic induction at a distance r (4 cm) is given as \(10^{-8}\) tesla. Now, we can find the magnetic induction at a distance 3r (12 cm) by multiplying the known induction by the ratio we found in Step 4: \[B_2 = B_1 \times \frac{1}{3} = (10^{-8}) \times \frac{1}{3}\]
06

Calculate B₂

Calculate the new magnetic induction: \[B_2 = 10^{-8} \times \frac{1}{3} = 3.33 \times 10^{-9}\mathrm{~tesla}\]
07

Choose the correct answer

From our calculations, the magnetic induction at a distance of 12 cm from the same current-carrying wire is \(3.33 \times 10^{-9}\) tesla. Therefore, the correct answer is: (a) \(3.33 \times 10^{-9}\)

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Most popular questions from this chapter

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) \(\left(\mathrm{r}_{1} / \mathrm{r}_{2}\right)\)

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 magnetic field existing in a region is given by $\mathrm{B}^{-}=\mathrm{B}_{0}[1+(\mathrm{x} / \ell)] \mathrm{k} \wedge . \mathrm{A}\( square loop of side \)\ell$ and carrying current I is placed with edges (sides) parallel to \(\mathrm{X}-\mathrm{Y}\) axis. The magnitude of the net magnetic force experienced by the Loop is (a) \(2 \mathrm{~B}_{0} \overline{\mathrm{I} \ell}\) (b) \((1 / 2) B_{0} I \ell\) (c) \(\mathrm{B}_{\circ} \mathrm{I} \ell\) (d) BI\ell

A dip needle vibrates in the vertical plane perpendicular to the magnetic meridian. The time period of vibration is found to be \(2 \mathrm{sec}\). The same needle is then allowed to vibrate in the horizontal plane and the time period is again found to be \(2 \mathrm{sec}\). Then the angle of dip is (a) \(0^{\circ}\) (b) \(30^{\circ}\) (c) \(45^{\circ}\) (d) \(90^{\circ}\)

0: Due to 10 Amp of current flowing in a circular coil of \(10 \mathrm{~cm}\) radius, the mag. field produced at its centre is \(\pi \times 10^{-3}\) Tesla. The number of turns in the coil will be (a) 5000 (b) 100 (c) 50 (d) 25

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