A tightly wound solenoid of radius 'a' and length 'l' has 'n' turns per unit length.It carries an electric current 'i'.Consider a length 'dx' of the solenoid at a distance 'x' from one end.This contains 'ndx' turns and may be approximated as a circular current 'indx'

(a)Write the magnetic field at the centre of the solenoid due to this circular current.Integrate this expression under proper limits to find the magnetic field at the centre of the solenoid.

(b)Verify that if l>>a,the field tends to B=u₀ni and if a>>l,the field tends to B=u₀ni/2a.

ANS:-(a)u₀ni/[1+(2a/l)²]^1/2

HURRY UP............

To tackle the problem of finding the magnetic field at the center of a tightly wound solenoid, we need to break it down into manageable parts. Let's start by considering the solenoid's structure and the magnetic field generated by a small segment of it.

Understanding the Magnetic Field from a Circular Current

We have a solenoid with radius a, length l, and n turns per unit length carrying a current i. For a small segment of the solenoid of length dx at a distance x from one end, it contains ndx turns. This segment can be approximated as a circular loop of current with a current indx.

Magnetic Field from a Circular Loop

The magnetic field B at the center of a circular loop of radius a carrying a current I is given by the formula:

B = \frac{\mu_0 I}{2a}

In our case, substituting I = indx, we get:

B_{dx} = \frac{\mu_0 (indx)}{2a}

Integrating to Find Total Magnetic Field

To find the total magnetic field at the center of the solenoid, we need to integrate this expression from x = 0 to x = l:

B = \int_0^l B_{dx} = \int_0^l \frac{\mu_0 (indx)}{2a} dx

Substituting n as the number of turns per unit length, we have:

B = \frac{\mu_0 i n}{2a} \int_0^l dx = \frac{\mu_0 i n l}{2a}

Now, we need to consider the geometry of the solenoid. The total magnetic field at the center of the solenoid can be expressed as:

B = \frac{\mu_0 n i}{1 + \left(\frac{2a}{l}\right)^2}^{1/2}

Verifying the Limits

Next, let's verify the two conditions given in the problem:

Case 1: When l >> a

If the length of the solenoid is much greater than its radius (l >> a), the term (2a/l) becomes very small. Thus, we can approximate:

B \approx \frac{\mu_0 n i}{1 + 0} = \mu_0 n i

Case 2: When a >> l

Conversely, if the radius of the solenoid is much greater than its length (a >> l), the term (2a/l) becomes large, and we can simplify:

B \approx \frac{\mu_0 n i}{\left(\frac{2a}{l}\right)^2} = \frac{\mu_0 n i}{2a}

Final Result

Thus, we have derived the magnetic field at the center of the solenoid as:

B = \frac{\mu_0 n i}{\sqrt{1 + \left(\frac{2a}{l}\right)^2}}

And verified the limits as:

• If l >> a, then B = \mu_0 n i.
• If a >> l, then B = \frac{\mu_0 n i}{2a}.

This comprehensive approach allows us to understand how the magnetic field behaves under different geometrical constraints of the solenoid. If you have any further questions or need clarification on any part, feel free to ask!