Question 1 |
In the figure, the electric field E and the magnetic field B point to \mathrm{x} and \mathrm{z} directions, respectively, and have constant magnitudes. A positive charge 'q' is released from rest at the origin. Which of the following statement(s) is/ are true.


The charge will move in the direction of z with constant velocity. | |
The charge will al ways move on the y-z plane only. | |
The trajectory of the charge will be a cycle. | |
The charge will progress in the direction of y. |
Question 1 Explanation:
As per Answer key of IIT Official : MTA (Marks to All)
Given :
Given :
Question 2 |
The vector function expressed by
F = a_x(5y - k_1z) + a_y(3z + k_2x) + a_z(k_3y - 4x)
represents a conservative field, where a_x, a_y, a_z are unit vectors along x, y and z directions, respectively. The values of constants k_1, k_2, k_3 are given by:
F = a_x(5y - k_1z) + a_y(3z + k_2x) + a_z(k_3y - 4x)
represents a conservative field, where a_x, a_y, a_z are unit vectors along x, y and z directions, respectively. The values of constants k_1, k_2, k_3 are given by:
k_1=3, k_2=3, k_3=7 | |
k_1=3, k_2=8, k_3=5 | |
k_1=4, k_2=5, k_3=3 | |
k_1=0, k_2=0, k_3=0 |
Question 2 Explanation:
\bar{F}=(5y-k_{1}Z)\hat{i}+(3z+k_{2}x)\hat{j}+(k_{3}y-4x)\hat{k}
is conservative field
\bar{F} is irrotational,
\begin{aligned} \triangledown \times \bar{F}&=0\\ \begin{vmatrix} \hat{i} &\hat{j} &\hat{k} \\ \frac{\partial }{\partial x}&\frac{\partial }{\partial y} &\frac{\partial }{\partial z} \\ 5y-k_{1}z &3z-k_{2}x &k_{3}y -4x \end{vmatrix}&=0\\ \hat{i}(k_{3}-3)-\hat{j}(-4+k_{1})+\hat{k}(k_{2}-5)&=0\\ k_{3}-3&=0\\ 4-k_{1}&=0\\ k_{2}-5&=0\\ k_{1}&=4\\ k_{2}&=5\\ k_{3}&=3 \end{aligned}
\bar{F} is irrotational,
\begin{aligned} \triangledown \times \bar{F}&=0\\ \begin{vmatrix} \hat{i} &\hat{j} &\hat{k} \\ \frac{\partial }{\partial x}&\frac{\partial }{\partial y} &\frac{\partial }{\partial z} \\ 5y-k_{1}z &3z-k_{2}x &k_{3}y -4x \end{vmatrix}&=0\\ \hat{i}(k_{3}-3)-\hat{j}(-4+k_{1})+\hat{k}(k_{2}-5)&=0\\ k_{3}-3&=0\\ 4-k_{1}&=0\\ k_{2}-5&=0\\ k_{1}&=4\\ k_{2}&=5\\ k_{3}&=3 \end{aligned}
Question 3 |
The figures show diagrammatic representations of vector fields \vec{X},\vec{Y} and \vec{Z} respectively.
Which one of the following choices is true?


\bigtriangledown \cdot \vec{X}=0, \bigtriangledown \times \vec{Y}\neq 0,\bigtriangledown \times \vec{Z}=0 | |
\bigtriangledown \cdot \vec{X} \neq 0, \bigtriangledown \times \vec{Y}=0, \bigtriangledown \times \vec{Z} \neq 0 | |
\bigtriangledown \cdot \vec{X}\neq 0, \bigtriangledown \times \vec{Y}\neq 0, \bigtriangledown \times \vec{Z}\neq 0 | |
\bigtriangledown \cdot \vec{X}=0, \bigtriangledown \times \vec{Y}= 0, \bigtriangledown \times \vec{Z}=0 |
Question 3 Explanation:
\vec{X} is going away so \vec{\triangledown } \cdot \vec{X}\neq 0
\vec{Y} is moving circulator direction so \vec{\triangledown } \cdot \vec{Y}\neq 0
\vec{Z} has circular rotation so \vec{\triangledown } \cdot \vec{Z}\neq 0
\vec{Y} is moving circulator direction so \vec{\triangledown } \cdot \vec{Y}\neq 0
\vec{Z} has circular rotation so \vec{\triangledown } \cdot \vec{Z}\neq 0
Question 4 |
The line integral of the vector field F = 5xz \hat{i}+ (3x^{2} + 2y) \hat{j} + x^{2}z\hat{k} along a path from (0,0,0) to (1,1,1) parametrized by (t, t^{2}, t) is _____.
4.41 | |
2.26 | |
6.56 | |
8.34 |
Question 4 Explanation:
\begin{aligned} E &=5xZ\bar{i}+(3x^2 +2y)\bar{j}+x^2z\bar{k} \\ &=\int _C \bar{F}\bar{d}r \\ &= \int _C 5xzdx +(3x^2+2y)dy +x^2zdz\\ x&=t, \; y=t^2,\; z=t,\; t=0 \; to \; 1 \\ dx&=dt \\ dy &=2tdt,\; dz =dt\\ &=\int_{0}^{1} 5t^2dt+(3t^2+2t^2)2tdt+t^3dt\\ &= \int_{0}^{1} (5t^2+11t^3)dt\\ &=\left [ \frac{5t^3}{3}+\frac{11t^4}{4} \right ]_0^1=\frac{5}{3}+\frac{11}{4}=4.41 \end{aligned}
Question 5 |
In cylindrical coordinate system, the potential produced by a uniform ring charge is given by \psi = f(r, z) , where f is a continuous function of r and z. Let \vec{E} be the resulting electric field. Then the magnitude of \bigtriangledown \times \vec{E}
increases with r. | |
is 0. | |
is 3. | |
decreases with z. |
Question 5 Explanation:
V is given as static field in time invariant.
Hence, \triangledown \times E=0
Hence, \triangledown \times E=0
There are 5 questions to complete.