Animated Solution for Physics - Magnetic Effects of Current: A long solenoid has $200$ turns per cm and carries a current $I$. The magnetic field at its centre is $6.28\times 10^{-2}{\text{ Wb/m}}^2$. Another long solenoid has $100$ turns per cm and it carries a current $I/3$. The value of the magnetic field at its centre is

A Ratio Approach

Imagine you are working in a physics lab with two long solenoids. The first one is tightly wound, boasting $200$ turns per centimeter, and carries a steady current $I$. The second one is a bit more relaxed, with only $100$ turns per centimeter, and carries a smaller current of $I/3$. We are given the magnetic field at the center of the first solenoid and tasked with finding the magnetic field at the center of the second one.

To solve this, we need to recall the fundamental formula for the magnetic field $B$ at the center of a long solenoid. The equation is elegantly simple:

Here, ${\mu }_0$ is the permeability of free space, $n$ is the number of turns per unit length, and $I$ is the current flowing through the wire. This equation tells us that the magnetic field is directly proportional to both the turn density and the current.

We could calculate everything from scratch, but there is a much smarter way: taking a ratio. Let's write down the equations for both solenoids:

For Solenoid 1: For Solenoid 2:

By dividing the second equation by the first, the constant ${\mu }_0$ beautifully cancels out, leaving us with a clean relationship:

A quick tip: Notice that we don't even need to convert the turn density from ${\text{cm}}^{-1}$ to ${\text{m}}^{-1}$. Because we are taking a ratio, any conversion factor would appear in both the numerator and the denominator and simply cancel out. This saves time and reduces the chance of silly calculation errors!

Now, let's plug in the values given in the problem: - $n_1=200{\text{ cm}}^{-1}$ - $n_2=100{\text{ cm}}^{-1}$ - $I_1=I$ - $I_2=I/3$

Substituting these into our ratio equation:

Simplifying the fractions, we get:

This means the magnetic field in the second solenoid is exactly one-sixth of the magnetic field in the first solenoid.

Finally, we substitute the known value of $B_1=6.28\times 10^{-2}{\text{ Wb/m}}^2$:

Rounding to two decimal places to match our options, we get $1.05\times 10^{-2}{\text{ Wb/m}}^2$. This perfectly matches option (a).