Question 11:
easyCharge $q_2$ is at the centre of a circular path with radius $r$. Work done in carrying charge $q_1$, once around this equipotential path, would be (1994)
A circular path around a point charge is an equipotential surface because the distance $r$ from the central charge $q_2$ is constant. Since the potential difference between any two points on this path is zero, the work done $W = q\Delta V$ is zero.
Question 12:
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 13:
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 14:
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 15:
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 16:
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 17:
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 18:
easy1. A square loop of side $1 m$ and resistance $1 \Omega$ is placed in a magnetic field of $0.5 T$. If the plane of loop of perpendicular to the direction of a magnetic field, the magnetic flux through the loop is: (2022)
Area of the square loop is $A = 1 \times 1 = 1 m^2$.
Since the plane of the loop is perpendicular to the magnetic field, the normal to the loop is parallel to the field, so $\theta = 0^\circ$.
Magnetic flux is $\Phi = B A \cos(0^\circ) = 0.5 \times 1 \times 1 = 0.5 Wb$.
Question 19:
easy2. The magnetic flux linked with a coil (in Wb) is given by the equation $\phi = 5t^2 + 3t + 16$. The magnitude of induced emf in the coil at the fourth second will be: (2020-Covid)
Magnitude of induced emf is $e = |\frac{d\phi}{dt}|$.
Differentiating flux with respect to time: $\frac{d\phi}{dt} = 10t + 3$.
At $t = 4 s$, $e = 10(4) + 3 = 43 V$.
Question 20:
easy3. A 800 turn coil of effective area $0.05 m^2$ is kept perpendicular to a magnetic field of $5 \times 10^{-5} T$. When the plane of the coil is rotated by $90^\circ$ around any of its coplanar axis in $0.1 s$, the emf induced in the coil will be: (2019)
Initial flux is $\Phi_1 = N B A \cos(0^\circ) = 800 \times (5 \times 10^{-5}) \times 0.05 = 2 \times 10^{-3} Wb$.
After rotating by $90^\circ$, $\Phi_2 = N B A \cos(90^\circ) = 0$.
Magnitude of induced emf is $e = |\frac{\Delta \Phi}{\Delta t}| = \frac{2 \times 10^{-3} - 0}{0.1} = 0.02 V$.