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Chapter 16: Problem 43

When the number of turns in a coil is doubled without any change in the length of the coil, its self-inductance becomes (A) Four times (B) Doubled (C) Halved (D) Squared

Short Answer

Expert verified
The self-inductance of a coil becomes four times its initial value when the number of turns is doubled without any change in the length of the coil. The correct answer is (A) Four times.

Step by step solution

01

Self-Inductance Formula

The formula for self-inductance of a coil is given by: \(L = \mu_0 \frac{N^2 A}{l}\), where \(L\) is the self-inductance, \(\mu_0\) is the permeability of free space, \(N\) is the number of turns, \(A\) is the cross-sectional area, and \(l\) is the length of the coil. #Step 2: Determine self-inductance when the number of turns is doubled#
02

New Self-Inductance with Doubled Number of Turns

Now let's determine the new self-inductance when we double the number of turns, which means \(N \) will be replaced with \(2N\): \(L' = \mu_0 \frac{(2N)^2 A}{l}\) #Step 3: Compare the new self-inductance with the original self-inductance#
03

Comparing Self-Inductance Values

Divide the new self-inductance (\(L'\)) by the original self-inductance (\(L\)): \(\frac{L'}{L} = \frac{\mu_0 \frac{(2N)^2 A}{l}}{\mu_0 \frac{N^2 A}{l}}\) #Step 4: Simplify the ratio to find the change in self-inductance#
04

Simplify the Ratio

After canceling out the common terms and simplifying, \(\frac{L'}{L} = \frac{(2N)^2}{N^2} = \frac{4N^2}{N^2} = 4\) The self-induction becomes four times its initial value when the number of turns is doubled, without any change in the length of the coil. So, the correct answer is: (A) Four times

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Most popular questions from this chapter

A conducting rod \(P Q\) of length \(L=1.0 \mathrm{~m}\) is moving with a uniform speed \(v=2.0 \mathrm{~m} / \mathrm{s}\) in a uniform magnetic field \(B=4.0 \mathrm{~T}\) direction into the paper. A capacitor of capacity \(C=10 \mu \mathrm{F}\) is connected as shown in Fig. \(16.44\). Then Fig. \(16.44\) (A) \(q_{A}=+80 \mu \mathrm{C}\) and \(q_{B}=-80 \mu \mathrm{C}\). (B) \(q_{A}=-80 \mu \mathrm{C}\) and \(q_{B}=+80 \mu \mathrm{C}\). (C) \(q_{A}=0=q_{B}\). (D) charge stored in the capacitor increases exponentially with time.

Figure \(16.66\) shows four rods having \(\lambda=0.5 \Omega / \mathrm{m}\) resistance per unit length. The arrangement is kept in a magnetic field of constant magnitude \(B=2 \mathrm{~T}\) and directed perpendicular to the plane of the figure and directed inwards. Initially, the rods form a square of side length \(\ell=15 \mathrm{~m}\) as shown. Now each wire starts moving with constant velocity \(v=5 \mathrm{~m} / \mathrm{s}\) towards opposite wire. Find the force required in newton on each wire to keep its velocity constant at \(t=1 \mathrm{~s}\).

Two straight long conductors \(A O B\) and \(C O D\) are perpendicular to each other and carry currents \(I_{1}\) and \(I_{2}\), respectively. The magnitude of the magnetic induction at a point \(P\) at a distance \(a\) from the point \(O\) in a direction perpendicular to the plane \(A B C D\) is (A) \(\frac{\mu_{0}}{2 \pi a}\left(I_{1}+I_{2}\right)\) (B) \(\frac{\mu_{0}}{2 \pi a}\left(I_{1}-I_{2}\right)\) (C) \(\frac{\mu_{0}}{2 \pi a}\left(I_{1}^{2}+I_{2}^{2}\right)^{1 / 2}\) (D) \(\frac{\mu_{0}}{2 \pi a}\left(\frac{I_{1} I_{2}}{I_{1}+I_{2}}\right)\)

A metal conductor of length \(1 \mathrm{~m}\) rotates vertically about one of its ends at angular velocity 5 radians per second. If the horizontal component of earth's magnetic field is \(0.2 \times 10^{-4} \mathrm{~T}\), then the EMF developed between the two ends of the conductor is (A) \(5 \mathrm{mV}\) (B) \(50 \mu \mathrm{V}\) (C) \(5 \mu \mathrm{V}\) (D) \(50 \mathrm{mV}\)

The magnetic field perpendicular to the plane of conducting ring of radius \(r\) changes at the rate \(\frac{d B}{d t}=\alpha\). Then (A) EMF induced in the ring is \(\pi r^{2} \alpha\). (B) EMF induced in the ring is \(2 \pi r \alpha\). (C) The potential difference between diametrically opposite points on the ring is half of induced EMF. (D) All points on the ring are at same potential.
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