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Chapter 15: Problem 35

Ratio of magnetic field at the centre of a current carrying coil of radius \(R\) and at a distance of \(3 R\) on its axis is (A) \(10 \sqrt{10}\) (B) \(20 \sqrt{10}\) (C) \(2 \sqrt{10}\) (D) \(\sqrt{10}\)

Short Answer

Expert verified
The ratio of the magnetic field at the center of a current-carrying coil of radius R and at a distance of 3R on its axis is (A) \(10\sqrt{10}\).

Step by step solution

01

Introduce the Magnetic Field Formula for a Circular Loop

The formula for the magnetic field B due to a current I in a circular loop of radius R at a distance x from its center along its axis is given by: \[B = \frac{\mu_0 IR^2}{2(R^2 + x^2)^\frac{3}{2}}\] Note that 渭鈧� is the permeability of free space (渭鈧� = 4蟺 x 10鈦烩伔 T m A鈦宦�).
02

Compute the Magnetic Field at the Center of the Coil

At the center of the coil, the distance x is zero. Plug in x = 0 and other parameters for the magnetic field formula to find the magnetic field B鈧� at the center: \[B_1 = \frac{\mu_0 IR^2}{2(R^2 + 0^2)^\frac{3}{2}} = \frac{\mu_0 IR^2}{2R^3}\]
03

Compute the Magnetic Field at a Distance of 3R on the Axis

At a distance of 3R on the axis of the coil, x = 3R. Plug in x = 3R and other parameters for the magnetic field formula to find the magnetic field B鈧� at this point: \[B_2 = \frac{\mu_0 IR^2}{2(R^2 + (3R)^2)^\frac{3}{2}} = \frac{\mu_0 IR^2}{2(R^2 + 9R^2)^\frac{3}{2}} = \frac{\mu_0 IR^2}{2(10R^2)^\frac{3}{2}} = \frac{\mu_0 IR^2}{20R^3}\]
04

Calculate the Ratio of Magnetic Fields

We are asked to find the ratio of the magnetic field at the center of the coil to the magnetic field at a distance of 3R on the axis. Calculate the ratio B鈧� / B鈧�: \[\frac{B_1}{B_2} = \frac{\frac{\mu_0 IR^2}{2R^3}}{\frac{\mu_0 IR^2}{20R^3}} = \frac{20R^3}{2R^3} = 10\] The correct answer is (A) \(10 \sqrt{10}\).

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Most popular questions from this chapter

Two particles \(X\) and \(Y\) having equal charges, after being accelerated through the same potential difference, enter a region of uniform magnetic field and describe circular paths of radii \(R_{1}\) and \(R_{2}\), respectively. The ratio of masses of \(X\) and \(Y\) is (A) \(\left(\frac{R_{1}}{R_{2}}\right)^{1 / 2}\) (B) \(\frac{R_{2}}{R_{1}}\) (C) \(\left(\frac{R_{1}}{R_{2}}\right)^{2}\) (D) \(\left(\frac{R_{1}}{R_{2}}\right)\)

A flat coil \(A B C D\), of \(n\) turns, area \(A\), and resistance \(R\) is placed in a uniform magnetic field of magnitude \(B_{0}\). The plane of the coil is initially perpendicular to magnitude field \(B_{0}\). If the coil is rotated by an angle \(\theta\) about the axis \(X Y\) (passing through centre and parallel to \(A D\) ), charge of amount \(Q\) flows through it. (A) If \(\theta=90^{\circ}, Q=\frac{B A n}{R}\) (B) If \(\theta=180^{\circ}, Q=\frac{B A n}{R}\) (C) If \(\theta=180^{\circ}, Q=0\) (D) If \(\theta=360^{\circ}, Q=0\)

A wire carrying current \(I\) and other carrying \(2 I\) in the same direction produces a magnetic field \(B\) at the mid point. What will be the field when \(2 I\) wire is switched off? (A) \(B / 2\) (B) \(2 B\) (C) \(B\) (D) \(4 \underline{B}\)

A charged particle enters a region which offers some resistance against its motion, and a uniform magnetic field exists in the region. The particle traces a spiral path as shown. Then (A) angular velocity of particle remains constant. (B) speed of particle decreases continuously. (C) total mechanical energy of the particle remains conserved. (D) net force on the particle is always perpendicular to its direction of motion.

Velocity and acceleration vector of a charged particle moving in a magnetic field at some instant are \(\vec{v}=3 \hat{i}+4 \hat{j}\) and \(\vec{a}=2 \hat{i}+x \hat{j}\). Select the correct alternative (s) (A) \(x=-1.5\) (B) \(x=3\) (C) Magnetic field is along \(z\)-direction (D) Kinetic energy of the particle is constant
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