Question

Two concentric loops $A$ and $B$ of same radius $2\pi~\text{cm$ are placed at right angles to each other. If the currents flowing through $A$ and $B$ are $3~\text{A}$ and $4~\text{A}$ respectively, then the net magnetic field at their common center is}

• A. $0.75 \times 10^{-5}~\text{T}$
• B. $25 \times 10^{-5}~\text{T}$
• C. $5 \times 10^{-5}~\text{T}$
• D. $2.5 \times 10^{-5}~\text{T}$

Hint:

Use vector addition for perpendicular fields: $B_{net} = \sqrt{B_1^2 + B_2^2}$.

Answer 1

The Correct Option is C

Field due to loop: $B = \dfrac{\mu_0 I}{2R}$
$R = 2\pi~\text{cm} = 2\pi \times 10^{-2}~\text{m}$
$B_A = \dfrac{4\pi \times 10^{-7} \cdot 3}{2 \cdot 2\pi \cdot 10^{-2}} = \dfrac{6 \times 10^{-7}}{4 \cdot 10^{-2}} = 1.5 \times 10^{-5}~\text{T}$
$B_B = \dfrac{4\pi \times 10^{-7} \cdot 4}{2 \cdot 2\pi \cdot 10^{-2}} = 2 \times 10^{-5}~\text{T}$
Net $B = \sqrt{(1.5)^2 + (2)^2} \times 10^{-5} = \sqrt{2.25 + 4} \cdot 10^{-5} = \sqrt{6.25} \cdot 10^{-5} = 2.5 \cdot 10^{-5}~\text{T}$