Question 211:
easyThe angle between the electric lines of force and the equipotential surface is: (2002)
Electric lines of force are always perpendicular to equipotential surfaces. This ensures that the component of the electric field along the equipotential surface is zero, meaning no work is done in moving a charge along it. Thus, the angle is $90^{\circ}$.
Question 212:
easyIn a certain region of space with volume $0.2 \text{ m}^3$, the electric potential is found to be $5 \text{ V}$ throughout. The magnitude of electric field in this region is: (2020)
The relationship between electric field and potential is $E = -\frac{dV}{dr}$. Since the electric potential $V$ is constant at $5 \text{ V}$ throughout the region, the gradient $\frac{dV}{dr}$ is zero. Therefore, the magnitude of the electric field is zero.
Question 213:
easyIf potential (in volts) in a region is expressed as $V(x, y, z) = 6xy – y + 2yz$, the electric field (in N/C) at point $(1, 1, 0)$ is: (2015)
Electric field $E = -\nabla V = -(\frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j} + \frac{\partial V}{\partial z}\hat{k})$. $\frac{\partial V}{\partial x} = 6y$, $\frac{\partial V}{\partial y} = 6x - 1 + 2z$, $\frac{\partial V}{\partial z} = 2y$. At point $(1, 1, 0)$, $E_x = -6(1) = -6$, $E_y = -(6(1) - 1 + 0) = -5$, $E_z = -2(1) = -2$. Thus, $E = -(6\hat{i} + 5\hat{j} + 2\hat{k})$.
Question 214:
easyIn a region, the potential is represented by $V(x, y, z) = 6x – 8xy – 8y + 6yz$, where $V$ is in volts and $x, y, z$ are in meters. The electric force experienced by a charge of $2 \text{ coulomb}$ situated at point $(1, 1, 1)$ is: (2014)
$E = -\nabla V$. Partial derivatives at $(1,1,1)$: $E_x = -(6 - 8y) = 2$, $E_y = -(-8x - 8 + 6z) = 10$, $E_z = -(6y) = -6$. Magnitude $|E| = \sqrt{2^2 + 10^2 + (-6)^2} = \sqrt{140} = 2\sqrt{35} \text{ N/C}$. Force $F = qE = 2 \times 2\sqrt{35} = 4\sqrt{35} \text{ N}$.
Question 215:
1. Across a metallic conductor of non-uniform cross section a constant potential difference is applied. The quantity which remains constant along the conductor is: (2015)
In a steady state, the current flowing through any cross-section of a conductor remains constant regardless of the cross-sectional area. Therefore, current remains constant throughout the metallic conductor.
Question 216:
easy2. A flow of $10^{7}$ electrons per second in a conducting wire constitutes a current of (1994)
Current $I = \frac{q}{t} = \frac{ne}{t}$. Given $\frac{n}{t} = 10^{7} \text{ s}^{-1}$. Thus, $I = 10^{7} \times 1.6 \times 10^{-19} = 1.6 \times 10^{-12} \text{ A}$.
Question 217:
easy3. The velocity of charge carriers of current (about $1 \text{ ampere}$) in a metal under normal conditions is of the order of (1991)
The drift velocity of electrons in a typical metallic conductor under normal conditions is extremely small, typically on the order of $10^{-4} \text{ m/s}$ or a fraction of a $\text{mm/sec}$.
Question 218:
An electric bulb is rated $60\text{ W}$, $220\text{ V}$. The resistance of its filament is (1994)
The resistance is given by $R = \frac{V^2}{P}$. Substituting the values, $R = \frac{(220)^2}{60} = \frac{48400}{60} \approx 806.67\Omega$, which rounds to $807\Omega$.
Question 219:
38. When a charged particle moving with velocity $\vec{v}$ is subjected to a magnetic field of induction $\vec{B}$, the force on it is non-zero. This implies that: (2006)
Force is $F = qvB\sin\theta$. For a non-zero force, $\sin\theta \neq 0$, meaning the angle cannot be $0^\circ$ or $180^\circ$, allowing any other value.
Question 220:
10. At a point A on the earth’s surface the angle of dip, $\delta = +25^{\circ}$. At a point B on the earth’s surface the angle of dip, $\delta = -25^{\circ}$. We can interpret that: (2019)
The angle of dip is conventionally taken as positive in the northern hemisphere and negative in the southern hemisphere. Therefore, A is in the northern hemisphere and B is in the southern hemisphere.