Question

Solve this one please.

Answer 1

The magnetic field of the galvanometer is 4.5 × 10⁵ Nm⁻¹A⁻¹

Explanation:

Given the twist in the coil $\tehta=45^\circ$

Current in the coil $I=1$ μA

Restoring torque per unit twist $k=10^{-4}$ Nm/deg

Number of turns $N=1000$

Area of the coil $A=10^{-5}$ m²

We know that, the relation between the above quantities and magnetic field B in a galvanometer is given by

$\boxed{I=(\frac{k}{NBA})\theta}$

$\implies B=\frac{k\theta}{INA}$

$\implies B=\frac{10^{-4}\times 45^\circ}{1\times 10^{-6}\times 1000\times 10^{-5}}$

$\implies B=4.5\times 10^5$ N/m-A

Hope this answer is helpful,

Know More:

Q:  Moving coil galvanometer (MCG) has 10 turns each of length  12 cm and breadth 8 cm. The coil of M.C. G. carries a current  of 125uA and is kept perpendicular to the uniform magnetic  field of induction 10 T. The twist constant of phosphor bronze  fibre is 12x10 Nm/degree. Calculate the deflection  produced​.

Click Here: https://brainly.in/question/15634817

Q: The coil area of a galvanometer is 25 × $10^{-4}$. It consists of 150 turns of a wire and is in a magnetic field of 0.15 T. The restoring torque constant of the suspension fibre is $10^{-6}$ Nm per degree. Assuming the magnetic field to be radial, calculate the maximum current that can be measured by the galvanometer, if the scale can accommodate 30° deflection.

Click Here: https://brainly.in/question/7920960

Answer 2

Given :

Twist , θ  = 45°

Current , I = 10 ^ (-6) A

Area of the coil A = 10 ^ (-5) m²

Number of turns , N = 1000

Restoring torque per unit twist ,  k = 10 ^ (-4) Nm/deg

To Find :

The magnetic field of the magnet of the galvanometer .

Solution :

We know , current in the moving coil galvanometer is given as

I = ( K * θ ) / ( N * B * A )

=> B = ( K * θ ) / ( N * I * A )

=> B = [ 10 ^ (-4) * 45 ] / [ 1000 * 10 ^ (-6) * 10 ^ (-5) ]

=> B = 45 * 10 ^ (4)

=> B = 4.5 * 10 ^ (4) N / Am

The magnetic field of the magnet of the galvanometer is 4.5 * 10 ^ (4) N / Am .