Explanation:

Assertion is correct by Faraday's law of electromagnetic induction. The Lorentz force \(\vec{F}=q(\vec{v} \times \vec{B})\) is the mechanism behind this induction, acting on the free electrons within the copper wire as it moves in the magnetic field. Both Assertion and Reason are correct and Reason is the correct explanation of Assertion. So correct option is (1).

Explanation:

Explanation:

Induced electric field at point \(P\)
\(E=\dfrac{R}{2} \dfrac{d B}{d t}\) towards right
Acceleration of electron
\(a=\dfrac{e E}{m}=\dfrac{e R}{2 m} \dfrac{d B}{d t}\) towards left

Explanation:

Method - 1
Consider a small element \(d x\) on the rod


The emf across the element is \(\mathrm{d} \varepsilon=\mathrm{E} \cos \theta \mathrm{dx}\)
\(\cos \theta=\dfrac{\mathrm{R}}{\mathrm{r}}, \operatorname{Tan} \theta=\dfrac{x}{\mathrm{R}} \Rightarrow x=\mathrm{R}\) Tan \(\theta\)
\(\mathrm{d} x=R \sec ^{2} \theta \mathrm{d} \theta\)
\(\mathrm{E}=\dfrac{\mathrm{R}^{2}}{2 \mathrm{r}} \dfrac{\mathrm{dB}}{\mathrm{dt}}=\dfrac{\mathrm{R}^{2} \alpha}{2 \mathrm{r}}\)
\(\begin{aligned}& \mathrm{d} \varepsilon=\dfrac{\mathrm{R}^{2} \alpha}{2 \mathrm{r}}\left(\dfrac{\mathrm{R}}{\mathrm{r}}\right) \mathrm{Rsec}^{2} \theta \mathrm{d} \theta \\& \mathrm{d} \varepsilon=\dfrac{\mathrm{R}^{4} \alpha}{2 \mathrm{r}^{2}} \sec ^{2} \theta \mathrm{d} \theta \\& \mathrm{d} \varepsilon=\dfrac{\mathrm{R}^{4} \alpha}{2 \mathrm{r}^{2}}\left(\dfrac{\mathrm{r}}{\mathrm{R}}\right)^{2} \mathrm{~d} \theta \\& \varepsilon=\dfrac{\mathrm{R}^{2} \alpha}{2} \int_{0}^{\pi / 2} \mathrm{~d} \theta=\dfrac{\mathrm{R}^{2} \alpha \pi}{4}\end{aligned}\)
Method - II
Join the two rods of the semi-infinite rod to the centre. With conducting wires the total flux through \(OAB\), where \(B\) is at the other end of the rod, is
\(\phi = \left( {\frac{\pi }{4}} \right){R^2}\;B\)
As one-fourth of the circular area is present inside the triangle \(OAB\)
\({\varepsilon _{{\text{ind }}}} = \frac{{d\phi }}{{dt}} = \frac{{\pi {R^2}}}{4}\frac{{dB}}{{dt}} = \frac{{\pi {R^2}\alpha }}{4}\)

Explanation:

Assertion is correct by Faraday's law of electromagnetic induction. The Lorentz force \(\vec{F}=q(\vec{v} \times \vec{B})\) is the mechanism behind this induction, acting on the free electrons within the copper wire as it moves in the magnetic field. Both Assertion and Reason are correct and Reason is the correct explanation of Assertion. So correct option is (1).

Explanation:

Explanation:

Induced electric field at point \(P\)
\(E=\dfrac{R}{2} \dfrac{d B}{d t}\) towards right
Acceleration of electron
\(a=\dfrac{e E}{m}=\dfrac{e R}{2 m} \dfrac{d B}{d t}\) towards left

Explanation:

Method - 1
Consider a small element \(d x\) on the rod


The emf across the element is \(\mathrm{d} \varepsilon=\mathrm{E} \cos \theta \mathrm{dx}\)
\(\cos \theta=\dfrac{\mathrm{R}}{\mathrm{r}}, \operatorname{Tan} \theta=\dfrac{x}{\mathrm{R}} \Rightarrow x=\mathrm{R}\) Tan \(\theta\)
\(\mathrm{d} x=R \sec ^{2} \theta \mathrm{d} \theta\)
\(\mathrm{E}=\dfrac{\mathrm{R}^{2}}{2 \mathrm{r}} \dfrac{\mathrm{dB}}{\mathrm{dt}}=\dfrac{\mathrm{R}^{2} \alpha}{2 \mathrm{r}}\)
\(\begin{aligned}& \mathrm{d} \varepsilon=\dfrac{\mathrm{R}^{2} \alpha}{2 \mathrm{r}}\left(\dfrac{\mathrm{R}}{\mathrm{r}}\right) \mathrm{Rsec}^{2} \theta \mathrm{d} \theta \\& \mathrm{d} \varepsilon=\dfrac{\mathrm{R}^{4} \alpha}{2 \mathrm{r}^{2}} \sec ^{2} \theta \mathrm{d} \theta \\& \mathrm{d} \varepsilon=\dfrac{\mathrm{R}^{4} \alpha}{2 \mathrm{r}^{2}}\left(\dfrac{\mathrm{r}}{\mathrm{R}}\right)^{2} \mathrm{~d} \theta \\& \varepsilon=\dfrac{\mathrm{R}^{2} \alpha}{2} \int_{0}^{\pi / 2} \mathrm{~d} \theta=\dfrac{\mathrm{R}^{2} \alpha \pi}{4}\end{aligned}\)
Method - II
Join the two rods of the semi-infinite rod to the centre. With conducting wires the total flux through \(OAB\), where \(B\) is at the other end of the rod, is
\(\phi = \left( {\frac{\pi }{4}} \right){R^2}\;B\)
As one-fourth of the circular area is present inside the triangle \(OAB\)
\({\varepsilon _{{\text{ind }}}} = \frac{{d\phi }}{{dt}} = \frac{{\pi {R^2}}}{4}\frac{{dB}}{{dt}} = \frac{{\pi {R^2}\alpha }}{4}\)

Explanation:

Assertion is correct by Faraday's law of electromagnetic induction. The Lorentz force \(\vec{F}=q(\vec{v} \times \vec{B})\) is the mechanism behind this induction, acting on the free electrons within the copper wire as it moves in the magnetic field. Both Assertion and Reason are correct and Reason is the correct explanation of Assertion. So correct option is (1).

Explanation:

Explanation:

Induced electric field at point \(P\)
\(E=\dfrac{R}{2} \dfrac{d B}{d t}\) towards right
Acceleration of electron
\(a=\dfrac{e E}{m}=\dfrac{e R}{2 m} \dfrac{d B}{d t}\) towards left

Explanation:

Method - 1
Consider a small element \(d x\) on the rod


The emf across the element is \(\mathrm{d} \varepsilon=\mathrm{E} \cos \theta \mathrm{dx}\)
\(\cos \theta=\dfrac{\mathrm{R}}{\mathrm{r}}, \operatorname{Tan} \theta=\dfrac{x}{\mathrm{R}} \Rightarrow x=\mathrm{R}\) Tan \(\theta\)
\(\mathrm{d} x=R \sec ^{2} \theta \mathrm{d} \theta\)
\(\mathrm{E}=\dfrac{\mathrm{R}^{2}}{2 \mathrm{r}} \dfrac{\mathrm{dB}}{\mathrm{dt}}=\dfrac{\mathrm{R}^{2} \alpha}{2 \mathrm{r}}\)
\(\begin{aligned}& \mathrm{d} \varepsilon=\dfrac{\mathrm{R}^{2} \alpha}{2 \mathrm{r}}\left(\dfrac{\mathrm{R}}{\mathrm{r}}\right) \mathrm{Rsec}^{2} \theta \mathrm{d} \theta \\& \mathrm{d} \varepsilon=\dfrac{\mathrm{R}^{4} \alpha}{2 \mathrm{r}^{2}} \sec ^{2} \theta \mathrm{d} \theta \\& \mathrm{d} \varepsilon=\dfrac{\mathrm{R}^{4} \alpha}{2 \mathrm{r}^{2}}\left(\dfrac{\mathrm{r}}{\mathrm{R}}\right)^{2} \mathrm{~d} \theta \\& \varepsilon=\dfrac{\mathrm{R}^{2} \alpha}{2} \int_{0}^{\pi / 2} \mathrm{~d} \theta=\dfrac{\mathrm{R}^{2} \alpha \pi}{4}\end{aligned}\)
Method - II
Join the two rods of the semi-infinite rod to the centre. With conducting wires the total flux through \(OAB\), where \(B\) is at the other end of the rod, is
\(\phi = \left( {\frac{\pi }{4}} \right){R^2}\;B\)
As one-fourth of the circular area is present inside the triangle \(OAB\)
\({\varepsilon _{{\text{ind }}}} = \frac{{d\phi }}{{dt}} = \frac{{\pi {R^2}}}{4}\frac{{dB}}{{dt}} = \frac{{\pi {R^2}\alpha }}{4}\)

Explanation:

Assertion is correct by Faraday's law of electromagnetic induction. The Lorentz force \(\vec{F}=q(\vec{v} \times \vec{B})\) is the mechanism behind this induction, acting on the free electrons within the copper wire as it moves in the magnetic field. Both Assertion and Reason are correct and Reason is the correct explanation of Assertion. So correct option is (1).

Explanation:

Explanation:

Induced electric field at point \(P\)
\(E=\dfrac{R}{2} \dfrac{d B}{d t}\) towards right
Acceleration of electron
\(a=\dfrac{e E}{m}=\dfrac{e R}{2 m} \dfrac{d B}{d t}\) towards left

Explanation:

Method - 1
Consider a small element \(d x\) on the rod


The emf across the element is \(\mathrm{d} \varepsilon=\mathrm{E} \cos \theta \mathrm{dx}\)
\(\cos \theta=\dfrac{\mathrm{R}}{\mathrm{r}}, \operatorname{Tan} \theta=\dfrac{x}{\mathrm{R}} \Rightarrow x=\mathrm{R}\) Tan \(\theta\)
\(\mathrm{d} x=R \sec ^{2} \theta \mathrm{d} \theta\)
\(\mathrm{E}=\dfrac{\mathrm{R}^{2}}{2 \mathrm{r}} \dfrac{\mathrm{dB}}{\mathrm{dt}}=\dfrac{\mathrm{R}^{2} \alpha}{2 \mathrm{r}}\)
\(\begin{aligned}& \mathrm{d} \varepsilon=\dfrac{\mathrm{R}^{2} \alpha}{2 \mathrm{r}}\left(\dfrac{\mathrm{R}}{\mathrm{r}}\right) \mathrm{Rsec}^{2} \theta \mathrm{d} \theta \\& \mathrm{d} \varepsilon=\dfrac{\mathrm{R}^{4} \alpha}{2 \mathrm{r}^{2}} \sec ^{2} \theta \mathrm{d} \theta \\& \mathrm{d} \varepsilon=\dfrac{\mathrm{R}^{4} \alpha}{2 \mathrm{r}^{2}}\left(\dfrac{\mathrm{r}}{\mathrm{R}}\right)^{2} \mathrm{~d} \theta \\& \varepsilon=\dfrac{\mathrm{R}^{2} \alpha}{2} \int_{0}^{\pi / 2} \mathrm{~d} \theta=\dfrac{\mathrm{R}^{2} \alpha \pi}{4}\end{aligned}\)
Method - II
Join the two rods of the semi-infinite rod to the centre. With conducting wires the total flux through \(OAB\), where \(B\) is at the other end of the rod, is
\(\phi = \left( {\frac{\pi }{4}} \right){R^2}\;B\)
As one-fourth of the circular area is present inside the triangle \(OAB\)
\({\varepsilon _{{\text{ind }}}} = \frac{{d\phi }}{{dt}} = \frac{{\pi {R^2}}}{4}\frac{{dB}}{{dt}} = \frac{{\pi {R^2}\alpha }}{4}\)