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Chapter 13: Problem 1887

Two concentric co-planar circular Loops of radii \(\mathrm{r}_{1}\) and \(\mathrm{r}_{2}\) carry currents of respectively \(\mathrm{I}_{1}\) and \(\mathrm{I}_{2}\) in opposite directions. The magnetic induction at the centre of the Loops is half that due to \(\mathrm{I}_{1}\) alone at the centre. If \(\mathrm{r}_{2}=2 \mathrm{r}_{1}\) the value of $\left(\mathrm{I}_{2} / \mathrm{I}_{1}\right)$ is (a) 2 (b) \(1 / 2\) (c) \(1 / 4\) (d) 1

Short Answer

Expert verified
The value of \(\frac{I_2}{I_1}\) is 1, which corresponds to the answer (d) 1.

Step by step solution

01

Calculate the magnetic field at the center of loop 1

Since loop 1 carries \(I_1\) current and has a radius of \(r_1\), the magnetic field at the center of loop 1 can be calculated by: \[B_1 = \frac{\mu_0 I_1}{2r_1}\]
02

Calculate the magnetic field at the center of loop 2

Since loop 2 carries \(I_2\) current and has a radius of \(r_2\), the magnetic field at the center of loop 2 can be calculated by: \[B_2 = \frac{\mu_0 I_2}{2r_2}\] Given that the currents are in opposite directions, we need to subtract the magnetic fields.
03

Given condition

According to the given condition, the magnetic induction at the center of the loops(B) is half that due to \(I_1\) alone at the center. Therefore, \[B = B_1 - B_2 = \frac{1}{2} B_1\]
04

Substitute the values of B_1 and B_2

Now, replace the expressions of \(B_1\) and \(B_2\) that we found in Step 1 and Step 2 into the equation and solve for \(\frac{I_2}{I_1}\): \[\frac{\mu_0 I_1}{2r_1} - \frac{\mu_0 I_2}{4r_1} = \frac{1}{2} \frac{\mu_0 I_1}{2r_1}\]
05

Solve for I_2/I_1

Simplify the equation and solve for \(\frac{I_2}{I_1}\): \[\frac{I_1 - I_2 / 2}{2} = \frac{I_1}{4}\] \[I_1 - \frac{I_2}{2} = \frac{I_1}{2}\] \[I_1 - \frac{I_1}{2} = \frac{I_2}{2}\] \[\frac{I_2}{I_1} = 1\] The value of \(\frac{I_2}{I_1}\) is 1, which corresponds to the answer (d) 1.

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Most popular questions from this chapter

A long wire carr1es a steady current. It is bent into a circle of one turn and the magnetic field at the centre of the coil is \(\mathrm{B}\). It is then bent into a circular Loop of n turns. The magnetic field at the centre of the coil for same current will be. (a) \(\mathrm{nB}\) (b) \(\mathrm{n}^{2} \mathrm{~B}\) (c) \(2 \mathrm{nB}\) (d) \(2 \mathrm{n}^{2} \mathrm{~B}\)

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At a distance of \(10 \mathrm{~cm}\) from a long straight wire carrying current, the magnetic field is \(4 \times 10^{-2}\). At the distance of $40 \mathrm{~cm}$, the magnetic field will be Tesla. (a) \(1 \times 10^{-2}\) (b) \(2 \times \overline{10^{-2}}\) (c) \(8 \times 10^{-2}\) (d) \(16 \times 10^{-2}\)
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