Question 1 |
Two identical copper wires W1 and W2, placed in parallel as shown in the figure, carry currents I and 2I, respectively, in opposite directions. If the two wires are separated by a distance of 4r, then the magnitude of the magnetic field \vec{B} between the wires at a distance r from W1 is
\frac{\mu _0 I}{6 \pi r} | |
\frac{6 \mu _0 I}{5 \pi r} | |
\frac{5 \mu _0 I}{6 \pi r} | |
\frac{{\mu _0}^2 I^2}{2 \pi r^2} |
Question 1 Explanation:
Magnetic flux density (\vec{B}) at r distance due to infinite line carrying current I is |\vec{B}|=\frac{\mu_{0} I}{2 \pi \rho} .
\vec{B} at r distance due to W_{1} wire
=\left|\vec{B}_{1}\right|=\frac{\mu_{0} I}{2 \pi r}\qquad \ldots(i)
\vec{B} at 3r distance due to W_{2} wire
=\left|\vec{B}_{2}\right|=\frac{\mu_{0}(2 I)}{2 \pi(3 r)}\qquad \ldots(ii)
From right hand thumb rule, \vec{B} due to both lines add in between conductors.
\begin{aligned} {So,}\qquad |\vec{B}|&=\left|\vec{B}_{1}\right|+\left|\vec{B}_{2}\right|\\ \therefore \qquad |\vec{B}|&=\frac{\mu_{0} I}{2 \pi r}+\frac{2 \mu_{0} I}{6 \pi r}=\frac{5 \mu_{0} I}{6 \pi r} \end{aligned}
Question 2 |
In the table shown, List I and List II, respectively, contain terms appearing on the left-hand side and the right-hand side of Maxwell's equations (in their standard form). Match the left-hand side with the corresponding right-hand side.
1-P, 2-R, 3-Q, 4-S | |
1-Q, 2-R, 3-P, 4-S | |
1-Q, 2-S, 3-P, 4-R | |
1-R, 2-Q, 3-S, 4-P |
Question 2 Explanation:
\begin{aligned} \nabla \cdot \vec{D} &=\rho_{v} \\ \nabla \times \vec{E} &=-\frac{\partial \vec{B}}{\partial t} \\ \nabla \cdot \vec{B} &=0 \\ \nabla \times \vec{H} &=\vec{J}+\frac{\partial \vec{D}}{\partial t} \end{aligned}
Question 3 |
What is the electric flux (\int \vec{E}\cdot d\hat{a}) through a quarter-cylinder of height H (as shown in the figure) due to an infinitely long line charge along the axis of the cylinder with a charge density of Q?
\frac{HQ}{\varepsilon _0} | |
\frac{HQ}{4\varepsilon _0} | |
\frac{H\varepsilon _0}{4Q} | |
\frac{4H}{Q\varepsilon _0} |
Question 3 Explanation:
Electric field intensity (\vec{E}) at '\rho' distance due to infinite long line having line charge density Q is
\begin{aligned} \vec{E} &=\frac{Q}{2 \pi \varepsilon_{0} \rho} \hat{a}_{\rho} \\ \int \vec{E} \cdot \vec{da} &=\iint \frac{Q}{2 \pi \varepsilon_{0} \rho} \hat{a}_{p} \cdot \rho d \phi d z \hat{a}_{\rho} \\ &=\frac{Q}{2 \pi \varepsilon_{0}} \int_{<\pi / 2>} d \phi \int_{z=0}^{H} d z \\ &=\frac{Q}{2 \pi \varepsilon_{0}}\left(\frac{\pi}{2}\right) H=\frac{H Q}{4 \varepsilon_{0}} \end{aligned}
\begin{aligned} \vec{E} &=\frac{Q}{2 \pi \varepsilon_{0} \rho} \hat{a}_{\rho} \\ \int \vec{E} \cdot \vec{da} &=\iint \frac{Q}{2 \pi \varepsilon_{0} \rho} \hat{a}_{p} \cdot \rho d \phi d z \hat{a}_{\rho} \\ &=\frac{Q}{2 \pi \varepsilon_{0}} \int_{<\pi / 2>} d \phi \int_{z=0}^{H} d z \\ &=\frac{Q}{2 \pi \varepsilon_{0}}\left(\frac{\pi}{2}\right) H=\frac{H Q}{4 \varepsilon_{0}} \end{aligned}
Question 4 |
An electron (q_1) is moving in free space with velocity 10^{5} m/s towards a stationary electron (q_2) far away. The closest distance that this moving electron gets to the stationary electron before the repulsive force diverts its path is ___________ \times 10^{-8}m.
[Given, mass of electron m=9.11 \times 10^{-31} kg, charge of electron e=-1.6 \times 10^{-19}C, and permittivity \varepsilon_{0}=(1/36\pi)\times 10^{-9}F/m]
[Given, mass of electron m=9.11 \times 10^{-31} kg, charge of electron e=-1.6 \times 10^{-19}C, and permittivity \varepsilon_{0}=(1/36\pi)\times 10^{-9}F/m]
4 | |
5 | |
6 | |
7 |
Question 4 Explanation:
r is the distance at which kinetic energy of q_{1} becomes zero [(because kinetic energy (KE) is converted into potential energy (PE)].
When q_{1} reaches 'r', it starts diverting.
Kinetic energy, K E=\frac{1}{2} m v^{2} and work done in moving q_{1} charge to distance 'r' is
\begin{aligned} q_{1} v_{2}&=q_{1} \frac{q_{2}}{4 \pi \varepsilon_{0} r} \\ &\qquad\left(q_{1}=q_{2}=-1.6 \times 10^{-19} \mathrm{C}\right)\\ Now, \quad \frac{1}{2} m v^{2}&=\frac{q_{1} q_{2}}{4 \pi \varepsilon_{0} r} \\ \end{aligned}
\Rightarrow r=\frac{\left(2 \times-1.6 \times 10^{-19}\right) \times\left(-1.6 \times 10^{-19}\right)}{4 \pi \times \frac{10^{-9}}{36 \pi} \times 9.11 \times 10^{-31} \times\left(10^{5}\right)^{2}}
\simeq 5.06 \times 10^{-8} \mathrm{m}
Question 5 |
Two conducting spheres S1 and S2 of radii a and b (b>a) respectively, are placed far apart and connected by a long, thin conducting wire, as shown in the figure.
For some charge placed on this structure, the potential and surface electric field on S1 are V_{a} and E_{a}, and that on S2 are V_{b} and E_{b}, respectively, which of the following is CORRECT?
For some charge placed on this structure, the potential and surface electric field on S1 are V_{a} and E_{a}, and that on S2 are V_{b} and E_{b}, respectively, which of the following is CORRECT?
V_{a}=V_{b} \; and \; E_{a}\lt E_{b} | |
V_{a}\gt V_{b} \; and \; E_{a} \gt E_{b} | |
V_{a}=V_{b} \; and \; E_{a} \gt E_{b} | |
V_{a}\gt V_{b} \; and \; E_{a}=E_{b} |
Question 5 Explanation:
When charge is placed on this structure equilibrium is established such that be spheres are at same potential i.e.
V_{a}=V_{b}
\begin{array}{c} V_{a}=V_{b} \\ \text { So, } \frac{Q_{a}}{4 \pi \epsilon_{o} a}=\frac{Q_{b}}{4 \pi \epsilon_{o} b} \end{array}
\frac{Q_{b}}{Q_{a}}=\frac{b}{a}
Now, surface electric fields.
\frac{E_{a}}{E_{b}}=\left[\frac{Q_{a} / 4 \pi \varepsilon_{o} a^{2}}{Q_{b} / 4 \pi \varepsilon_{o} b^{2}}\right]=\frac{Q_{a} \times b^{2}}{Q_{b} \times a^{2}}=\frac{b}{a}>1
So, E_{a}>E_{b}
Question 6 |
Consider the charge profile shown in the figure. The resultant potential distribution is best described by
A | |
B | |
C | |
D |
Question 6 Explanation:
Applying Poisson's equations
\nabla^{2} V=\frac{\partial^{2} V}{\partial x^{2}}=-\frac{\rho_{v}}{\epsilon}=K
Constant charge density
\begin{aligned} \frac{\partial V}{\partial x} &=-K x+K^{\prime} \\ V &=\frac{-K x^{2}}{2}+K^{\prime} x+K^{\prime} \end{aligned}
Towards positive x or negative side.
It is a second order parabolic increase.
Due to symmetry of + and - charges K" = 0 is expected with V = O at centre and graph passing through origin.
Beyond x \gt 0 \text{ or } x \lt b, \; E = 0 due to capacitive nature of + and - charges
V=-\int 0 \cdot \overrightarrow{d l}=\text { Constant }
This constant is same V at x=a or x=b
\nabla^{2} V=\frac{\partial^{2} V}{\partial x^{2}}=-\frac{\rho_{v}}{\epsilon}=K
Constant charge density
\begin{aligned} \frac{\partial V}{\partial x} &=-K x+K^{\prime} \\ V &=\frac{-K x^{2}}{2}+K^{\prime} x+K^{\prime} \end{aligned}
Towards positive x or negative side.
It is a second order parabolic increase.
Due to symmetry of + and - charges K" = 0 is expected with V = O at centre and graph passing through origin.
Beyond x \gt 0 \text{ or } x \lt b, \; E = 0 due to capacitive nature of + and - charges
V=-\int 0 \cdot \overrightarrow{d l}=\text { Constant }
This constant is same V at x=a or x=b
Question 7 |
A positive charge q is placed at x= 0 between two infinite metal plates placed at x =-d and at x = +d respectively. The metal plates lie in the yz plane.
The charge is at rest at t = 0, when a voltage +V is applied to the plate at -d and voltage -V is applied to the plate at x=+d . Assume that the quantity of the charge q is small enough that it does not perturb the field set up by the metal plates. The time that the charge q takes to reach the right plate is proportional to
The charge is at rest at t = 0, when a voltage +V is applied to the plate at -d and voltage -V is applied to the plate at x=+d . Assume that the quantity of the charge q is small enough that it does not perturb the field set up by the metal plates. The time that the charge q takes to reach the right plate is proportional to
d/V | |
\sqrt{d}/V | |
d/ \sqrt{V} | |
\sqrt{d/V} |
Question 7 Explanation:
Velocity being free velocity,
\begin{aligned} \frac{1}{2} m v^{2} &=q V \\ v &=\frac{d}{t}=\sqrt{\frac{2 q V}{m}} \\ \Rightarrow t \propto \frac{d}{\sqrt{V}} \end{aligned}
\begin{aligned} \frac{1}{2} m v^{2} &=q V \\ v &=\frac{d}{t}=\sqrt{\frac{2 q V}{m}} \\ \Rightarrow t \propto \frac{d}{\sqrt{V}} \end{aligned}
Question 8 |
The parallel-plate capacitor shown in the figure has movable plates. The capacitor is charged so that the energy stored in it is E when the plate separation is d. The capacitor is then isolated electrically and the plates are moved such that the plate separation becomes 2d.
At this new plate separation, what is the energy stored in the capacitor, neglecting fringing effects?
At this new plate separation, what is the energy stored in the capacitor, neglecting fringing effects?
2E | |
\sqrt{2}E | |
E | |
E/2 |
Question 8 Explanation:
\begin{array}{l} \text { Let, } E=E_{1} \text { , Energy } E_{1}=\frac{Q_{1}^{2}}{2 C_{1}} \\ \text { Electrically isolated } \Rightarrow Q_{2}=Q_{1} \\ \begin{aligned} d_{2}=2 d_{1} \Rightarrow \;& C_{2}=\frac{C_{1}}{2} \\ E_{2} &=\frac{Q_{2}^{2}}{2 C_{2}}=\frac{Q_{1}^{2}}{\frac{2 C_{1}}{2}}=2\left(\frac{Q_{1}^{2}}{2 C_{1}}\right) \\ &=2 E_{1}=2 E \end{aligned} \end{array}
Question 9 |
A uniform and constant magnetic field B=\hat{z}B exists in the \hat{z} direction in vacuum. A particle of mass m with a small charge q is introduced into this region with an initial velocity v=\hat{x}v_{x}+\hat{z}v_{z}. Given that B, m, q, v_{x} and v_{z} are all non-zero, which one of the following describes the eventual trajectory of the particle?
Helical motion in the \hat{z} direction | |
Circular motion in the xy plane | |
Linear motion in the \hat{z} direction | |
Linear motion in the \hat{x} direction |
Question 9 Explanation:
\mathrm{Ba}_{\mathrm{z}} magnetic field
v_{x} a_{x}+v_{z} a_{z} velocity
\begin{aligned} F&=Q(v \times B) \text{ by Lorentz's law}\\ &=Q\left(v_{x} a_{x}+v_{z} a_{z}\right) \times B a_{z} \\ F_{y} &=Q v_{x} \cdot B\left(-a_{y}\right) \end{aligned}
This results in a circular path in the XY plane with v_{z} a_{z} component causing a linear path.
Both result in a helical path along Z axis.
v_{x} a_{x}+v_{z} a_{z} velocity
\begin{aligned} F&=Q(v \times B) \text{ by Lorentz's law}\\ &=Q\left(v_{x} a_{x}+v_{z} a_{z}\right) \times B a_{z} \\ F_{y} &=Q v_{x} \cdot B\left(-a_{y}\right) \end{aligned}
This results in a circular path in the XY plane with v_{z} a_{z} component causing a linear path.
Both result in a helical path along Z axis.
Question 10 |
The current density in a medium is given by
\vec{j}=\frac{400 sin\theta }{2\pi (r^{2})+4}\hat{a}_{r}Am^{-2}
The total current and the average current density flowing through the portion of a spherical surface r = 0.8 m, \frac{\pi }{12}\leq \theta \leq \frac{\pi }{4},0\leq \phi \leq 2\pi are given, respectively, by
\vec{j}=\frac{400 sin\theta }{2\pi (r^{2})+4}\hat{a}_{r}Am^{-2}
The total current and the average current density flowing through the portion of a spherical surface r = 0.8 m, \frac{\pi }{12}\leq \theta \leq \frac{\pi }{4},0\leq \phi \leq 2\pi are given, respectively, by
15.09 A, 12.86 Am^{-2} | |
18.73 A, 13.65 Am^{-2} | |
12.86 A, 9.23 Am^{-2} | |
10.28 A, 7.56 Am^{-2} |
Question 10 Explanation:
\begin{aligned} I &=\int \overrightarrow{J} \cdot \overrightarrow{d s} \\ &=\int_{\theta=\frac{\pi}{12}}^{\pi / 4} \int_{\phi=0}^{2 \pi} \frac{400 \sin \theta}{2 \pi\left(r^{2}+4\right)} r^{2} \sin \theta d \theta d \phi \\ &=\left.\frac{400}{2 \pi\left(r^{2}+4\right)} \cdot r^{2} \cdot \phi\right|_{0} ^{2 \pi} \int_{\pi / 12}^{\pi / 4} \sin ^{2} \theta d \theta \\ &=\frac{400 r^{2}}{\left(r^{2}+4\right)} \int_{\pi / 12}^{\pi / 4}\left(\frac{1-\cos 2 \theta}{2}\right) d \theta \\ &=\frac{400 \cdot r^{2}}{\left(r^{2}+4\right)}\left(\frac{\left(\frac{\pi}{4}-\frac{\pi}{12}\right)}{2}-\left(\frac{\sin 2 \theta}{4}\right)_{\pi / 12}^{\pi / 4}\right) \\ &=\left.\frac{400 \cdot r^{2}}{\left(r^{2}+4\right)}\left(\frac{\pi}{12}-\left(\frac{1-1 / 2}{4}\right)\right)\right|_{r=0.8} \\ &=\frac{400 \times 0.8 \times 0.8}{4.64} \times 0.13=7.17 \mathrm{Amp} \end{aligned}
\begin{array}{l} \text { Total area }=\int d s=\iint r^{2} \sin \theta d \theta d \phi \\ =r^{2} \int_{0=\frac{\pi}{12}}^{\pi / 4} \sin \theta d \theta \cdot 2 \pi \\ =\left.r^{2} \cdot 2 \pi \cdot 0.259\right|_{r=0.8} \\ =0.8^{2} \times 0.5 \times 2 \pi \times \frac{1}{4} \\ =1.041 \mathrm{m}^{2} \\ \text { Average current }=\frac{7.17}{1.041}=6.88 \mathrm{A} / \mathrm{m}^{2} \end{array}
Note : GATE key mentioned (MTA - marks to all)
\begin{array}{l} \text { Total area }=\int d s=\iint r^{2} \sin \theta d \theta d \phi \\ =r^{2} \int_{0=\frac{\pi}{12}}^{\pi / 4} \sin \theta d \theta \cdot 2 \pi \\ =\left.r^{2} \cdot 2 \pi \cdot 0.259\right|_{r=0.8} \\ =0.8^{2} \times 0.5 \times 2 \pi \times \frac{1}{4} \\ =1.041 \mathrm{m}^{2} \\ \text { Average current }=\frac{7.17}{1.041}=6.88 \mathrm{A} / \mathrm{m}^{2} \end{array}
Note : GATE key mentioned (MTA - marks to all)
There are 10 questions to complete.