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Chapter 15: Problem 151

If in a circular coil \(A\) of radius \(R\), current \(I\) is flowing and in another coil \(B\) of radius \(2 R\) a current \(2 I\) is flowing, then the ratio of the magnetic fields \(B_{A}\) and \(B_{B}\) produced by them will be (A) 1 (B) 2 (C) \(\frac{1}{2}\) (D) 4

Short Answer

Expert verified
The ratio of the magnetic fields $B_{A}$ and $B_{B}$ produced by the two coils is (A) 1.

Step by step solution

01

Recall the formula for the magnetic field of a circular coil

To find the magnetic field produced by a circular coil with a current I and a radius R, we use the formula: \[ B = \frac{\mu_0 I}{2 R} \] where B is the magnetic field, \(\mu_0\) is the permeability of free space (a constant), and R is the radius of the coil.
02

Calculate the magnetic field of coil A

We plug the given values for coil A, \(R = R\) and \(I = I\), into the formula to find the magnetic field produced by coil A: \[B_{A} = \frac{\mu_0 I}{2 R} \]
03

Calculate the magnetic field of coil B

We plug the given values for coil B, \(R = 2R\) and \(I = 2I\), into the formula to find the magnetic field produced by coil B: \[B_{B} = \frac{\mu_0 (2I)}{2 (2R)} \]
04

Simplify the magnetic field of coil B

Simplify the magnetic field formula for coil B by canceling out some terms: \[B_{B} = \frac{\mu_0 I}{2R} \]
05

Find the ratio of the magnetic fields

Now we will divide the magnetic field of coil A by the magnetic field of coil B: \[\frac{B_{A}}{B_{B}} = \frac{\frac{\mu_0 I}{2 R}}{\frac{\mu_0 I}{2 R}} \]
06

Simplify the ratio

Simplify the ratio of the magnetic fields by canceling out terms: \[\frac{B_{A}}{B_{B}} = \frac{\cancel{\frac{\mu_0 I}{2 R}}}{\cancel{\frac{\mu_0 I}{2 R}}} = 1\] So, the ratio of the magnetic fields produced by coil A and coil B is: (A) 1

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Most popular questions from this chapter

A current \(I\) flows along the length of an infinitely long, straight, thin- walled pipe. Then [2007] (A) the magnetic field at all points inside the pipe is the same, but not zero. (B) the magnetic field is zero only on the axis of the pipe. (C) the magnetic field is different at different points inside the pipe. (D) the magnetic field at any point inside the pipe is zero.

A circular current carrying loop of radius \(R\), carries a current \(i\). The magnetic field at a point on the axis of coil is \(\frac{1}{\sqrt{8}}\) times the value of magnetic field at the centre. Distance of point from centre is (A) \(\frac{R}{\sqrt{2}}\) (B) \(\frac{R}{\sqrt{3}}\) (C) \(R \sqrt{2}\) (D) \(R\)

Two long conductors, separated by a distance \(d\), carry current \(I_{1}\) and \(I_{2}\) in the same direction. They exert a force \(F\) on each other. Now the current in one of them is doubles and its direction is reversed. The distance is also increased to \(3 d\). The new value of the force between them is (A) \(-\frac{2 F}{3}\) (B) \(\frac{F}{3}\) (C) \(-2 F\) (D) \(-\frac{F}{3}\)

Earth's magnetic induction at a certain point is \(7 \times 10^{-5}\) \(\mathrm{Wb} / \mathrm{m}^{2}\). This field is to be annulled by the magnetic induction at the centre of a circular conducing loop \(5.0 \mathrm{~cm}\) in radius. The required current is (A) \(0.056 \mathrm{~A}\) (B) \(6.5 \mathrm{~A}\) (C) \(5.6 \mathrm{~A}\) (D) \(12.8 \mathrm{~A}\)

A current-carrying conductor is looped into a circle of radius \(10 \mathrm{~cm}\). The magnetic moment of the current loop becomes \(0.314 \mathrm{~A} / \mathrm{m}^{2}\). What is the current in the loop? (A) \(5 \mathrm{~A}\) (B) \(8 \mathrm{~A}\) (C) \(10 \mathrm{~A}\) (D) \(12 \mathrm{~A}\)
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