Question

The magnetic field at the centre of current carrying coil is B₀ If its radius is reduced to half keeping current the "same then magnetic field at its centre become:
1. B₀
2. 2B₀
3. 4B₀
4. \(\frac{B_0}{2}\)

Answer 1

Correct Answer - Option 2 : 2B₀

CONCEPT:

• The field around a current-carrying wire or around a magnetic in which the magnetic force can be experienced by another current-carrying wire or by another magnet is called a magnetic field.

• The magnetic field at the center of a circular loop of radius 'R' is given by

\(\Rightarrow B = \frac{\mu _{0} I}{2 \pi R}\)

Where I  = Current R = Radius

EXPLANATION :

Let B₀ be the initial magnetic field and B = New magnetic field after changing the radius 

• The magnetic field at the center of the circular loop before changing the radius is given by 

​\(\Rightarrow B_{0} = \frac{\mu _{0} I}{2 \pi R}\)​

After changing the radius into half the new magnetic field can be written as 

\(\Rightarrow B = \frac{\mu _{0} I}{2 \pi \frac{R}{2}}\)

\(\Rightarrow B = 2 \frac{\mu_{0} I }{2\pi R}\)

\(\Rightarrow B = 2 B_{0}\)

• Hence, option 2 is the answer