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Chapter 13: Problem 1985

When 2 Amp current is passed through a tangent galvanometer. It gives a deflection of \(30^{\circ} .\) For \(60^{\circ}\) deflection, the current must be (a) \(1 \mathrm{Amp}\) (b) \(2 \sqrt{3 a m p}\) (c) \(4 \mathrm{amp}\) (d) \(6 \mathrm{Amp}\)

Short Answer

Expert verified
The current required for a \(60^{\circ}\) deflection in a tangent galvanometer is \(2\sqrt{3}Amp\).

Step by step solution

01

Understand the tangent galvanometer

A tangent galvanometer is an instrument used to measure the current by the deflection of a magnetic needle. The current flowing through the coil of the galvanometer creates a magnetic field, causing the needle to deflect. The deflection angle is proportional to the strength of the current.
02

Derive the relationship between current and deflection angle

In a tangent galvanometer, the tangent of the deflection angle is directly proportional to the current. This can be represented as: \( \tan\theta=kI \) where: θ = deflection angle I = current k = a constant For our given problem, we have a deflection angle of 30° corresponding to a current of 2 Amps.
03

Determine the constant k

To find the value of k, we can use the initial data provided in the problem: \( \tan30^{\circ}= k(2) \) \( \frac{1}{\sqrt{3}}=2k \) Solve for k: \( k =\frac {1}{2\sqrt{3}} \)
04

Calculate the current for the 60° deflection

Now we can use the equation we derived earlier to find the current corresponding to a 60° deflection: \( \tan60^{\circ}= kI \) Use the value of k we found previously: \( \sqrt{3} = \frac{1}{2\sqrt{3}}I \) Solve for I: \( I = 2\sqrt{3}Amps \)
05

Choose the correct answer

The current for a 60° deflection is 2√3 Amp. The correct answer is: (b) \(2\sqrt{3}Amp\)

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Most popular questions from this chapter

A long horizontal wire " \(\mathrm{A}^{\prime \prime}\) carries a current of $50 \mathrm{Amp}$. It is rigidly fixed. Another small wire "B" is placed just above and parallel to " \(\mathrm{A}^{\prime \prime}\). The weight of wire- \(\mathrm{B}\) per unit length is \(75 \times 10^{-3}\) Newton/meter and carries a current of 25 Amp. Find the position of wire \(B\) from \(A\) so that wire \(B\) remains suspended due to magnetic repulsion. Also indicate the direction of current in \(B\) w.r.t. to \(A\).(a) \((1 / 2) \times 10^{-2} \mathrm{~m}\); in same direction (b) \((1 / 3) \times 10^{-2} \mathrm{~m}\); in mutually opposite direction (c) \((1 / 4) \times 10^{-2} \mathrm{~m}\); in same direction (d) \((1 / 5) \times 10^{-2} \mathrm{~m}\); in mutually opposite direction

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) $\sqrt{5}\left(\mu_{0} / 4 \pi\right)\left(\mathrm{M} / \mathrm{d}^{3}\right)$

If a magnet of pole strength \(\mathrm{m}\) is divided into four parts such that the length and width of each part is half that of initial one, then the pole strength of each part will be (a) \((\mathrm{m} / 4)\) (b) \((\mathrm{m} / 2)\) (c) \((\mathrm{m} / 8)\) (d) \(4 \mathrm{~m}\)

The time period of a freely suspended magnet is a 4 seconds. If it is broken in length into two equal parts and one part is suspended in the same way, then its time period will be (a) \(4 \mathrm{sec}\) (b) \(2 \mathrm{sec}\) (c) \(0.5 \mathrm{sec}\) (d) \(0.25 \mathrm{sec}\)

A particle of mass \(\mathrm{m}\) and charge q moves with a constant velocity v along the positive \(\mathrm{x}\) -direction. It enters a region containing a uniform magnetic field B directed along the negative \(z\) -direction, extending from \(x=a\) to \(x=b\). The minimum value of required so that the particle can just enter the region \(\mathrm{x}>\mathrm{b}\) is (a) \([(\mathrm{q} \mathrm{bB}) / \mathrm{m}]\) (b) \(q(b-a)(B / m)\) (c) \([(\mathrm{qaB}) / \mathrm{m}]\) (d) \(q(b+a)(B / 2 m)\)
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