Question

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

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

Step-by-step solution

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.

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.

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: \( k =\frac {1}{2\sqrt{3}} \)

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: \( I = 2\sqrt{3}Amps \)

Choose the correct answer

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