Question

How would the strength of magnetic field due to current-carrying loop be affected if the radius of loop is reduced to half of its original value?

Answer 1

Step1: Expression for the magnetic field due to the current carrying loop

The magnetic field due to the current carrying loop is given as:

$B=\frac{{\mu }_{o}i}{2R}$ ………… (1)

Here ${\mu }_o$ is the permeability of free space, i is the current, R is the radius

Step2: Expression for the magnetic field when the radius is reduced to half

If the radius of the loop is reduced to half then: $R\text{'}=\frac{R}{2}$

Using the equation(1):

$$\begin{gathered} B\text{'}=\frac{{\mu }_{o}i}{2R\text{'}}=\frac{{\mu }_{o}i}{2{\displaystyle \frac{R}{2}}} \\ B\text{'}=2\left(\frac{{\mu }_{o}i}{2R}\right) \\ B\text{'}=2B \end{gathered}$$

Step3: Conclusion from the above expression

Therefore, the strength of the magnetic field due to the current-carrying loop is doubled if the radius is reduced to half of its original value.