Calculate the change in magnetic field induction at the centre of a current carrying loop of radius R, if the radius of coil is reduced to half and current through it changes from I to 2I .

Answer

To solve the problem, we need to calculate the change in magnetic field induction at the center of a current-carrying loop when the radius is reduced to half and the current is doubled.

### Step-by-Step Solution:

1. **Understand the Formula for Magnetic Field**:
The magnetic field \( B \) at the center of a circular loop carrying current \( I \) and having radius \( R \) is given by the formula:
\[ B = \frac{\mu_0 I}{2R} \]
where \( \mu_0 \) is the permeability of free space.

2. **Calculate Initial Magnetic Field \( B_1 \)**:
For the initial loop with radius \( R \) and current \( I \):
\[ B_1 = \frac{\mu_0 I}{2R} \]

3. **Calculate New Magnetic Field \( B_2 \)**:
For the new loop with radius \( \frac{R}{2} \) and current \( 2I \):
\[ B_2 = \frac{\mu_0 (2I)}{2 \left(\frac{R}{2}\right)} = \frac{\mu_0 (2I)}{R} = \frac{2\mu_0 I}{R} \]

4. **Calculate the Change in Magnetic Field \( \Delta B \)**:
The change in magnetic field is given by:
\[ \Delta B = B_2 - B_1 \]
Substituting the values we calculated:
\[ \Delta B = \frac{2\mu_0 I}{R} - \frac{\mu_0 I}{2R} \]
To combine these fractions, we need a common denominator:
\[ \Delta B = \frac{2\mu_0 I}{R} - \frac{\mu_0 I}{2R} = \frac{4\mu_0 I}{2R} - \frac{\mu_0 I}{2R} = \frac{3\mu_0 I}{2R} \]

5. **Final Result**:
The change in magnetic field induction at the center of the loop is:
\[ \Delta B = \frac{3\mu_0 I}{2R} \]