Question

A conducting loop of resistance $10\Omega$ and area $3.5x{10}^{-3}{m}^2$ is placed in uniform and time varying magnetic field$B=0.4T\sin \left(50\pi t\right)$. The Charge passing through the loop in$t=0ms$ to $t=10ms$ is:

• A. $140\mu C$
• B. $70\mu C$
• C. $280\mu C$
• D. $100\mu C$

Answer 1

The correct option is A

$140\mu C$

Solution:

Step 1. Given data:

Time dependent magnetic field, $B\left(t\right)=0.4T\sin \left(50\pi t\right)$

Area of the wire, $A=3.5x{10}^{-3}{m}^2$
Resistance of the wire, $R=10\mathrm{\Omega }$
On comparing time dependent magnetic field with the standard form of $B=B_0\sin \left(\omega t\right)$, we get:
$B_0=0.4T$ and $\omega =50\pi$

Step 2. Calculating total charge passing through the loop:
We know that the Faraday’s Law states that:

The induced emf in the loop
$V=\frac{d\varphi }{dt}$ $\_\_\_\_\left(1\right)$
The magnetic flux, $\varphi$ is given as:
$$\begin{gathered} \varphi =B.A \\ \Rightarrow \varphi =B_0\sin \left(\omega t\right).A \end{gathered}$$

Putting value of $\varphi$ in equation $\left(1\right)$ , we get:
$$\begin{gathered} V=\frac{d}{dt}(B_0\sin (\omega t).A) \\ \Rightarrow V=A{B}_0\frac{d}{dt}\left(\sin \omega t\right) \end{gathered}$$
$\Rightarrow V=A{B}_0\omega \cos \left(\omega t\right)$
Since, the voltage is directly proportional to charge flow, we can write it as:
$\Rightarrow \frac{dQ}{dt}R=A{B}_0\omega \cos \left(\omega t\right)$ $\left[V=R\frac{dQ}{dt}\right]$

$\Rightarrow \frac{dQ}{dt}=\frac{A{B}_0}{R}\omega \cos \left(\omega t\right)$
To get the amount of charge flowing, we integrate this equation:
$\int dQ=\underset{0}{\overset{10ms}{\int }}\frac{A{B}_0\omega }{R}\cos \left(\omega t\right)dt$
Taking the constant outside of the integral and solving gives us:
$$\begin{gathered} \Rightarrow Q=\frac{A{B}_0\omega }{R}\underset{0}{\overset{10ms}{\int }}\cos \left(\omega t\right)dt \\ \Rightarrow Q=\frac{A{B}_0\omega }{R}{\left|\frac{\sin \left(\omega t\right)}{\omega }\right|}_0^{10ms} \end{gathered}$$
$$\begin{gathered} \Rightarrow Q=\frac{A{B}_0}{R}{\left|\sin \left(\omega t\right)\right|}_0^{10ms} \\ \Rightarrow Q=\frac{A{B}_0}{R}[\sin \left(\omega \times 10\times {10}^{-3}\right)-\sin \left(\omega \times 0\right)] \end{gathered}$$ $\left[1ms={10}^{-3}s\right]$ $\_\_\_\_\left(2\right)$
On substituting the values in equation $\left(2\right)$, we get:

$\Rightarrow Q=\frac{3.5\times {10}^{-3}\times 0.4}{10}[\sin \left(50\pi \right)\times 10\times {10}^{-3}]$
$$\begin{gathered} \Rightarrow Q=0.14\times {10}^{-3}\sin \left(\frac{50\pi }{100}\right) \\ \Rightarrow Q=0.14\times {10}^{-3}\sin \left(\frac{\pi }{2}\right) \\ \Rightarrow Q=0.14\times {10}^{-3}C \\ \Rightarrow Q=140\times {10}^{-6}C \end{gathered}$$

$\therefore Q=140\mu C$ $\left[1\mu C={10}^{-6}C\right]$
Hence, the correct answer is option (A).