Question

A light and smooth insulating rod $PQ$ of length $R$ is fixed as a chord in a circular region of radius $R$. $C$ is mid-point of the rod. Answer the following questions. Magnetic field $B=2t$ exists in the circular region.

• A. 3/4
• B. 3/2
• C. 4/3
• D. 2/3

Answer 1

Answer: (B)

3/2

Answer 2

The induced electric field magnitude in a circular region with a time-varying magnetic field $B(t)$ is given by $E(r) = \frac{r}{2} \frac{dB}{dt}$, where $r$ is the distance from the center. Given $B = 2t$, so $\frac{dB}{dt} = 2$. Thus, $E(r) = \frac{r}{2}(2) = r$.

Let the origin O be at the center of the circular region. The rod PQ is a chord of length $R$ in a circle of radius $R$. The distance $d$ from the center O to the chord PQ is found using the Pythagorean theorem: $d^2 + (\frac{R}{2})^2 = R^2$, which gives $d = \frac{\sqrt{3}}{2}R$. Let the chord PQ lie along the line $y = -d = -\frac{\sqrt{3}}{2}R$. The coordinates of P and C are $P(-\frac{R}{2}, -\frac{\sqrt{3}}{2}R)$ and $C(0, -\frac{\sqrt{3}}{2}R)$.

The induced electric field is tangential. If we assume the magnetic field is into the page and increasing, the induced electric field is clockwise. Thus, $\vec{E} = y\hat{i} - x\hat{j}$. The magnitude is $|\vec{E}| = \sqrt{y^2 + (-x)^2} = \sqrt{x^2+y^2} = r$.

The induced emf between P and C is $\mathcal{E}_{PC} = \int_P^C \vec{E} \cdot d\vec{l}$. The path from P to C is along the rod, so $y = -\frac{\sqrt{3}}{2}R$ and $d\vec{l} = dx\hat{i}$. $\mathcal{E}_{PC} = \int_{-R/2}^{0} (y\hat{i} - x\hat{j}) \cdot (dx\hat{i}) = \int_{-R/2}^{0} y dx$. Substituting $y = -\frac{\sqrt{3}}{2}R$: $\mathcal{E}_{PC} = \int_{-R/2}^{0} (-\frac{\sqrt{3}}{2}R) dx = (-\frac{\sqrt{3}}{2}R) [x]_{-R/2}^{0} = (-\frac{\sqrt{3}}{2}R) (0 - (-\frac{R}{2})) = (-\frac{\sqrt{3}}{2}R)(\frac{R}{2}) = -\frac{\sqrt{3}}{4}R^2$.

The magnitude of the induced emf is $|\mathcal{E}_{PC}| = \frac{\sqrt{3}}{4}R^2$.

The problem states the magnitude is $\frac{\sqrt{a}}{b}tR^2$. This implies that the induced emf is time-dependent. However, with $B=2t$, the calculated emf is time-independent. This suggests a potential typo in the question.

If we assume the magnetic field is $B = 2t^2$, then $\frac{dB}{dt} = 4t$. The induced electric field magnitude is $E(r) = \frac{r}{2} \frac{dB}{dt} = \frac{r}{2}(4t) = 2rt$. The induced emf between P and C would be: $\mathcal{E}_{PC} = \int_P^C \vec{E} \cdot d\vec{l} = \int_{-R/2}^{0} (y\hat{i} - x\hat{j}) \cdot (dx\hat{i})$ where $y = -\frac{\sqrt{3}}{2}R$ and $\vec{E} = 2rt \hat{t}$. The tangential component of $\vec{E}$ at a point $(x,y)$ is $E_t = E \cos \theta$, where $\theta$ is the angle between the tangential direction and the x-axis. In our coordinate system, the tangential direction is given by $\hat{t} = \frac{-x\hat{i} + y\hat{j}}{r}$. So $\vec{E} = E \hat{t} = 2rt \frac{-x\hat{i} + y\hat{j}}{r} = 2t(-x\hat{i} + y\hat{j})$. $\vec{E} \cdot d\vec{l} = (2t(-x\hat{i} + y\hat{j})) \cdot (dx\hat{i}) = -2tx dx$. $\mathcal{E}_{PC} = \int_{-R/2}^{0} -2tx dx = -2t [\frac{x^2}{2}]_{-R/2}^{0} = -2t (0 - \frac{(-R/2)^2}{2}) = -2t (-\frac{R^2}{8}) = \frac{1}{4}tR^2$. This is not matching the expected form.

Let's re-evaluate the induced electric field. For a magnetic field $\vec{B} = B(t)\hat{k}$, the induced electric field is $\vec{E} = E(r)\hat{\phi}$, where $\hat{\phi}$ is the azimuthal unit vector. $E(r) = \frac{r}{2} \frac{dB}{dt}$. If $B = 2t$, $E(r) = r$. If $B = 2t^2$, $E(r) = 2rt$. The rod PQ is along the line $y = -d = -\frac{\sqrt{3}}{2}R$, from $x = -R/2$ to $x=0$. The tangential direction $\hat{\phi}$ is perpendicular to the radius vector $\vec{r} = x\hat{i} + y\hat{j}$. The direction of the rod is along the x-axis, so $d\vec{l} = dx\hat{i}$. The tangential unit vector $\hat{\phi}$ at a point $(x,y)$ is $\frac{-y\hat{i} + x\hat{j}}{r}$. So, $\vec{E} = E(r) \hat{\phi} = E(r) \frac{-y\hat{i} + x\hat{j}}{r}$.

If $B=2t$, $E(r)=r$. $\vec{E} = r \frac{-y\hat{i} + x\hat{j}}{r} = -y\hat{i} + x\hat{j}$. $\vec{E} \cdot d\vec{l} = (-y\hat{i} + x\hat{j}) \cdot (dx\hat{i}) = -y dx$. $\mathcal{E}_{PC} = \int_{-R/2}^{0} -y dx = \int_{-R/2}^{0} -(-\frac{\sqrt{3}}{2}R) dx = \frac{\sqrt{3}}{2}R \int_{-R/2}^{0} dx = \frac{\sqrt{3}}{2}R [\frac{R}{2}] = \frac{\sqrt{3}}{4}R^2$. (Magnitude)

If $B=2t^2$, $E(r)=2rt$. $\vec{E} = 2rt \frac{-y\hat{i} + x\hat{j}}{r} = 2t(-y\hat{i} + x\hat{j})$. $\vec{E} \cdot d\vec{l} = (2t(-y\hat{i} + x\hat{j})) \cdot (dx\hat{i}) = -2ty dx$. $\mathcal{E}_{PC} = \int_{-R/2}^{0} -2ty dx = \int_{-R/2}^{0} -2t(-\frac{\sqrt{3}}{2}R) dx = \sqrt{3}tR \int_{-R/2}^{0} dx$. $\mathcal{E}_{PC} = \sqrt{3}tR [x]_{-R/2}^{0} = \sqrt{3}tR (0 - (-\frac{R}{2})) = \sqrt{3}tR (\frac{R}{2}) = \frac{\sqrt{3}}{2}tR^2$.

The magnitude of the induced emf is $\frac{\sqrt{3}}{2}tR^2$. Comparing this with the given form $\frac{\sqrt{a}}{b}tR^2$: $\frac{\sqrt{3}}{2} = \frac{\sqrt{a}}{b}$. This implies $\sqrt{a} = \sqrt{3}$ and $b=2$. So $a=3$ and $b=2$. The value of $\frac{a}{b} = \frac{3}{2}$. This assumes the magnetic field was $B=2t^2$. Given the options, this is the most plausible interpretation.