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  • A rectangular loop of sides a and b, has a resistance R and lies at a distance c from an infinite straight wire carrying current <img alt="\mathrm{I}_0" src="https://entrancecorner.oncodecogs.com/g

A rectangular loop of sides a and b, has a resistance R and lies at a distance c from an infinite straight wire carrying current \mathrm{I}_0. The current decreases to zero in time \tau

\mathrm{I(t)=I_0 \frac{(\tau-t)}{\tau}, 0<\mathrm{t}<\tau}

The charge flowing through the rectangular loop is

Option: 1

\mathrm{\mu_0 I_0 \ \tau}


Option: 2

\mathrm{\mu_0 I_0 \frac{a b}{c^2} \tau}

 


Option: 3

\mathrm{\frac{1}{2} \frac{\mu_0 b I_0}{\pi R} \operatorname{In}\left(\frac{c+a}{c}\right)}


Option: 4

\mathrm{\frac{\mu_0 I \tau}{R} \frac{b a}{c^2}}


Answers (1)

best_answer

Consider an infinitesimal element of length b. thickness dx at a distance x from the wire. Since \mathrm{\mathrm{d} \phi=\mathrm{BdA}}

\mathrm{\begin{array}{ll} \Rightarrow & d \phi=\frac{\mu_0 I}{2 \pi x}(b d x_ 0) \\ \Rightarrow & \phi=\int d \phi=\frac{\mu_0 I b}{2 \pi} \int_c^{c+a} \frac{d x}{x} \\ \Rightarrow & \phi=\frac{\mu_0 b I}{2 \pi} \log _e\left(\frac{c+a}{c}\right) \\ \Rightarrow & \phi=\frac{\mu_0 b}{2 \pi} \log _e\left(\frac{c+a}{c}\right)\left[I_0\left(1-\frac{t}{\tau}\right)\right]\left\{\because I=I_0\left(1-\frac{t}{\tau}\right)\right\} \end{array}}

Since

\begin{array}{llll} & \varepsilon=-\frac{d \phi}{d t} & \Rightarrow & \varepsilon=-\frac{\mu_0 b I_0}{2 \pi} \log _e\left(\frac{c+a}{c}\right)\left(-\frac{1}{\tau}\right) \\ \Rightarrow \quad & \varepsilon=\frac{\mu_0 b I_0}{2 \pi \tau} \log _e\left(\frac{c+a}{c}\right) & \Rightarrow & I=\frac{\varepsilon}{R}=\frac{\mu_0 b I_0}{2 \pi \tau R} \log _e\left(\frac{c+a}{c}\right) \\ \Rightarrow \quad & \frac{d q}{d t}=\frac{\mu_0 b l_0}{2 \pi \tau R} \log _e\left(\frac{c+a}{c}\right) & \Rightarrow & d q=\frac{\mu_0 b l_0}{2 \pi \tau R} \log _e\left(\frac{c+a}{c}\right) d t \\ \Rightarrow \quad & q=\frac{\mu_0 b l_0}{2 \pi \tau R} \log _e\left(\frac{c+a}{c}\right) \int_0^\tau d t & \Rightarrow & q=\frac{\mu_0 b I_0}{2 \pi \tau R} \log _e\left(\frac{c+a}{c}\right) \end{array}

Posted by

himanshu.meshram

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