Magnetic Effects Of Current Question 10
Question 10 - 29 January - Shift 1
The magnitude of magnetic induction at mid-point $O$ due to current arrangement as shown in Fig will be :
(1) $\frac{\mu_0 I}{2 \pi a}$
(2) 0
(3) $\frac{\mu_0 I}{4 \pi a}$
(4) $\frac{\mu_0 I}{\pi a}$
Show Answer
Answer: (4)
Solution:
Formula: Magnetic field due to circular loop
Magnetic field due to current in $BC$ and ET are outward at point ’ $O$ '
$B_0=\frac{\mu_0 i}{4 \pi r}+\frac{\mu_0 i}{4 \pi r}=\frac{\mu_0 i}{2 \pi r}=\frac{\mu_0 i}{\pi a}$