Q. 153.9( 22 Votes )

# A magnetic field

Answer :

Here, we have a particular value of No. of turns per unit Length and Current in the coil in order to obtain the given magnetic field.

The Required Magnetic field B = 100 G = 100 × 10^{–4}= 10^{–2} T

Maximum Number of turns per unit length, n = 1000/m

Maximum Current flowing in the coil, I = 15 A

Permeability of free space, μ_{0} = 4π × 10^{–4}TmA^{-1}

We know magnetic field for a solenoid is given by

B = 𝜇 �_{0}nI

Where,

B is the magnetic field

n is the Number of turns of the coil per unit length

I is the current in the coil

𝜇 _{�0} is the permittivity of free space

__Explanation: here we can change the number of turns per unit length n (maximum 1000 turns per m) and current I (maximum 15A) since__ 𝜇_{o} __is a constant and we have to obtain a fixed value of magnetic field B = 10 ^{-2}T, so we will alter values of n and I accordingly__

Using the formula B = 𝜇 �_{0}nI

We get, B/𝜇 �_{0} = nI

So,

= 7957.74 A/m ≈ 8000 A/m

If we take number of turns of coil N = 400,

If we take length of the coil L = 50 cm = 0.5m,

No. of turns per unit length, n = N/L= 400/0.5 = 800/m

(less than 1000/m)

If we take current I = 10 A

(less than 15 A),

We get nI = 8000 A/m

Area of Coil A = 10^{-3} m^{2} = πr^{2}, where r is radius of coil

We get

⇒

⇒ r = 0.017m

The values of N, L and I are within the proposed limits, Though there can be other values as well.

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