A current I flows through a circular loop. If the radius of the loop is reduced to half and the current is doubled, the magnetic field intensity at the center becomes:
Four times the original value
To determine the change in magnetic field intensity at the center of a circular loop when its radius is reduced to half and the current is doubled, we need to use the formula for the magnetic field intensity at the center of a current-carrying loop.
The magnetic field B at the center of a circular loop carrying a current I with radius r is given by:
B = \frac{\mu_0 I}{2r}
Here, \mu_0 is the permeability of free space (a constant).
Initially, the magnetic field intensity B_1 is:
B_1 = \frac{\mu_0 I}{2r}
When the radius of the loop is reduced to half and the current is doubled, the new radius becomes \frac{r}{2} and the new current becomes 2I.
The new magnetic field intensity B_2 can be calculated as:
B_2 = \frac{\mu_0 (2I)}{2(\frac{r}{2})}
Simplifying this expression, we get:
B_2 = \frac{\mu_0 \cdot 2I}{r}
Further simplification gives:
B_2 = 4 \times \frac{\mu_0 I}{2r}
Thus:
B_2 = 4 \cdot B_1
Hence, the magnetic field intensity at the center becomes four times the original value when the radius is halved and the current is doubled.
Therefore, the correct answer is: Four times the original value.
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