A circular coil of radius R having N number of turns carries a steady current I. The magnetic induction at the centre of the coil is 0·1 tesla. If the number of turns is doubled and the radius is halved, which one of the following will be the correct value for the magnetic induction at the centre of the coil?
0·05 tesla
0·2 tesla
0·4 tesla
0·8 tesla
Answer is Right!
Answer is Wrong!
This question was previously asked in
UPSC NDA-2 – 2018
So, $0.1 = (\mu_0 N I) / (2R)$.
In the new scenario, the number of turns is doubled ($N_2 = 2N$) and the radius is halved ($R_2 = R/2$), while the current remains steady ($I_2 = I$).
The new magnetic field $B_2$ is given by:
$B_2 = (\mu_0 N_2 I_2) / (2R_2) = (\mu_0 (2N) I) / (2(R/2)) = (\mu_0 2N I) / R$.
We can express $B_2$ in terms of $B_1$:
$B_2 = (\mu_0 2N I) / R = 4 \times (\mu_0 N I) / (2R) = 4 \times B_1$.
Substituting the value of $B_1$:
$B_2 = 4 \times 0.1$ T = 0.4 T.