Question

$\text{ABC}$ is a triangular frame made out of a metallic wire. $\text{BD}$ is a median of the triangular frame joining the vertex $\text{B}$ to the side $\text{AC}$, (Also made of the same metallic wire). It is known that $\text{AB}=6\text{cm, BC}=8\text{cm}$ and $\text{AC}=10\text{cm}$. The cross-sectional area of the wire is $1{\text{mm}}^2$ and its resistivity is $24n\Omega \text{m}$. The triangle $\text{ABC}$ lies in a cylindrical region of magnetic field such that the intersection of the surface of the cylinder with the plane containing the frame forms the circumcircle of triangle $\text{ABC}$. The magnetic field in the region varies at the rate of $0.263\text{T/s}$. What is the magnitude of the induced current in $\text{mA}$ in the median $\text{BD}$ of the frame?

• A. $5\text{mA}$
• B. $10\text{mA}$
• C. $15\text{mA}$
• D. $20\text{mA}$

Answer 1

The correct option is B $10\text{mA}$

From the above geometry we can find that,

$\text{BD}=5\text{cm}$

Since, $\text{BD}$ is the median of triangle $\text{ABC}$

$\therefore Area\left(\text{ABD}\right)=Area\left(\text{BDC}\right)=A\left(\text{let}\right)$

So, magnitude of induced emf in the loop $\text{ABD}$ or $\text{BDC}$ will be

$E=|-\frac{d\varphi }{dt}|=A\frac{dB}{dt}$

$E=(\frac{1}{4}\times 8\times 6\times {10}^{-4})\times 0.263=12\times 0.263\times {10}^{-4}$

It means we can assume that both parts of triangle is connected to a battery of emf $E.$
Let $R$ be the resistance per $\text{cm}$ length of the wire.


Now applying $\text{KVL}$ to loops $\text{ABD}$ and $\text{BDC}$

$-5R{I}_1-5R(I_1-I_2)-6R{I}_1+E=0$

$16I_1-5I_2=\frac{E}{R}.........\left(1\right)$

Similarly,

$13R{I}_2-5R(I_1-I_2)=E$

$18I_2-5I_1=\frac{E}{R}..........\left(2\right)$

Solving this we get

$I_1=\frac{23}{263}\frac{E}{R},I_2=\frac{21}{263}\frac{E}{R}$

As we know $R=\frac{\rho l}{a}$

Here, $l=1\text{cm};a=1{\text{mm}}^2$

$R=\frac{(24\times {10}^{-9})(1\times {10}^{-2})}{1\times {10}^{-6}}=24\times {10}^{-5}\Omega$

$\Rightarrow I_2-I_1=\frac{2}{263}\frac{E}{R}$

$=\frac{2}{263}\times \frac{12\times 0.236\times {10}^{-4}}{24\times {10}^{-5}}$

$\therefore I_2-I_1=0.01\text{A}=10\text{mA}$

Hence, option $\left(B\right)$ is correct.