Curl MCQ Quiz in বাংলা - Objective Question with Answer for Curl - বিনামূল্যে ডাউনলোড করুন [PDF]

Concept:

\(\nabla.(ϕ (\vec{r}))=(\nabla ϕ).\vec{r}+ϕ(\nabla.\vec{r})\)

Explanation:

If ϕ(x,y,z) is scalar homogeneous function og degree 4 and \(\rm \vec{r}=x \vec{i}+y \vec{j}+z \vec{k}\)

Then, \(\rm \vec{r}=(x,y,z)\) & \(\nabla(\phi)=(\frac{\partial{\phi}}{\partial {x}},\frac{\partial{\phi}}{\partial {y}},\frac{\partial{\phi}}{\partial {z}})\)

Now, \(\rm \nabla .(ϕ \vec{r})=\nabla(\phi x \hat{\mathbf{i}}+\phi y \hat{\mathbf{j}}+\phi z \hat{\mathbf{k}})\)

Use, \(\nabla.(ϕ (\vec{r}))=(\nabla ϕ).\vec{r}+ϕ(\nabla.\vec{r})\)

Since, \(\nabla \vec{r}=3\)

\(\nabla.(ϕ (\vec{r}))=(\frac{\partial{\phi}}{\partial {x}},\frac{\partial{\phi}}{\partial {y}},\frac{\partial{\phi}}{\partial {z}}).(x,y,z)+\phi(\nabla . \vec{r})\)

= \(\rm x \frac{\partial{\phi}}{\partial{x}}+y\frac{\partial{\phi}}{\partial{y}}+z\frac{\partial{\phi}}{\partial{z}}+3\phi\)

Given \(\rm \phi(x,y,z)\) is homogenous function of degree 4, \(\rm \phi(λ x,λ y,λ z)=λ^4 \phi(x,y,z)\)

\(\rm x \frac{\partial{\phi}}{\partial{x}}+y\frac{\partial{\phi}}{\partial{y}}+z\frac{\partial{\phi}}{\partial{z}}=4λ^3 \phi(x,y,z)\)

Now, let λ=1

\(\rm x \frac{\partial{\phi}}{\partial{x}}+y\frac{\partial{\phi}}{\partial{y}}+z\frac{\partial{\phi}}{\partial{z}}=4\phi\)

\(\rm \nabla .(ϕ \vec{r})=3\phi+4\phi=7\phi\)   

\(\begin{array}{l} \nabla \times \vec F = 0\\ \left| {\begin{array}{*{20}{c}} i&j&k\\ {\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\\ {2x + 3y + \lambda z}&{\beta x + 6y + z}&{7x - \omega y + 8z} \end{array}} \right| = 0\\ i\left( { - \omega - 1 } \right) - j\left( {7 - \lambda } \right) + \hat k\left( {\beta - 3} \right) = 0\\ \Rightarrow \omega + 1 = 0\\\Rightarrow \omega = - 1\\ \beta = 3,\lambda =7\\ \lambda \beta \omega = 7 \times 3 \times -1 = - 21 \end{array}\)

Concept:

Curl \(\vec F = \vec \nabla \times \vec F\)

\(\vec \nabla \left( {\frac{\partial }{{\partial x}}\hat i + \frac{\partial }{{\partial y}}\hat j + \frac{\partial }{{\partial z}}\hat k} \right)\)

\(Curl\;\vec F = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k}\\ {\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\\ {{F_x}}&{{F_y}}&{{F_z}} \end{array}} \right|\)

Calculation:

= î (2z + 2xY) – ĵ (0 - 0) + k̂ (-2yz-2x²y)

At (1, 1, 1)

⇒  î (2 +2) + 0 ĵ + (-1 -2)k̂

Concept:

(i) \(\vec{F}\) Irrotational if \(\nabla \times \vec{F}=0\)

(ii) \(\vec{F}\) is solenoidal \(\nabla . \vec{F}=0\)

(iii) \(\nabla. \frac{\vec r}{r^3} = \frac{1}{r^3}\nabla.{\vec r}-\frac{3}{r^4}(r.\nabla r)\)

Explanation:

\(\nabla .F=\nabla. \frac{\vec r}{r^3} = \frac{1}{r^3}\nabla.{\vec r}-\frac{3}{r^4}(r.\nabla r)\)

=\(\frac{1}{r^3}(1+1+1)-\frac{3}{r^4}(x \frac{\partial r}{\partial x}+y \frac{\partial r}{\partial y}+z \frac{\partial r}{\partial z})\)

As, \(r = \sqrt{x^2+y^2+z^2}\)

\(\frac{\partial{r}}{\partial{x}}=\frac{x}{\sqrt{(x^2+y^2+z^2)}}=\frac{x}{r}\)

Similarly, \(\frac{\partial{r}}{\partial{y}}=\frac{y}{r}\) , \(\frac{\partial{r}}{\partial{z}}=\frac{z}{r}\)

=\(\frac{3}{r^4}-\frac{3}{r^4}(\frac{x^2}{r}+\frac{y^2}{r}+\frac{z^2}{r})\)

=\(\frac{3}{r^4}-\frac{3}{r^4}(\frac{r^2}{r})\)

=\(\frac{3}{r^4}-\frac{3}{r^3}=0\)

Now, \(\nabla \times \vec{F}\) = \(\nabla \times (\frac {\vec{r}}{r^3})\) = \(\nabla \times (\frac {x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}}{r^3})\) 

\(\nabla \times (\frac {\vec{r}}{r^3})\) = \(\begin{bmatrix} ​​​​\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ \frac{\partial}{\partial{x}} & \frac{\partial}{\partial{y}} & \frac{\partial}{\partial{z}} \\ \frac{x}{r^3} & \frac{y}{r^3} & \frac{z}{r^3} \end{bmatrix} \)

\(\nabla \times \vec{F} = (\frac{\partial}{\partial {y}}(\frac{z}{r^3})-\frac{\partial}{\partial {z}}(\frac{y}{r^3}))\hat{\mathbf{i}}+(\frac{\partial}{\partial {z}}(\frac{x}{r^3})-\frac{\partial}{\partial {x}}(\frac{z}{r^3}))\hat{\mathbf{j}}+(\frac{\partial}{\partial {x}}(\frac{y}{r^3})-\frac{\partial}{\partial {y}}(\frac{x}{r^3}))\hat{\mathbf{k}}\)

Now, \(\frac{\partial}{\partial{y}}(\frac{z}{r^3})=\frac{\partial}{\partial{y}}(\frac{z}{(x^2+y^2+z^2)^{3/2}})\)

= \(\frac{-3yz}{(x^2+y^2+z^2)^{3/2}}\)

Similarly, \(\frac{\partial}{\partial{y}}(\frac{y}{r^3})=\frac{\partial}{\partial{y}}(\frac{y}{(x^2+y^2+z^2)^{3/2}})\)=\(\frac{-3yz}{(x^2+y^2+z^2)^{3/2}}\)

⇒  \(\frac{\partial}{\partial {y}}(\frac{z}{r^3})-\frac{\partial}{\partial {z}}(\frac{y}{r^3})=0\)

Similarly, \(\frac{\partial}{\partial {z}}(\frac{x}{r^3})-\frac{\partial}{\partial {x}}(\frac{z}{r^3})=0\) & \(\frac{\partial}{\partial {x}}(\frac{y}{r^3})-\frac{\partial}{\partial {y}}(\frac{x}{r^3})=0\)

⇒ \(\nabla \times \vec{F}\) =0 

Concept:

Let ϕ be a function of (x, y, z)

Then grad \(\left( ϕ \right)=\hat{i}\frac{\partial ϕ }{\partial x}+\hat{j}\frac{\partial ϕ }{\partial y}+\hat{k}\frac{\partial ϕ }{\partial z}\)

curl (grad (ϕ)) \(=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\ \frac{\partial ϕ }{\partial x} & \frac{\partial ϕ }{\partial y} & \frac{\partial ϕ }{\partial z} \\ \end{matrix} \right|\)

curl (grad (ϕ)) \(=0\hat{i}+0\hat{j}+0\hat{k}\)

Explanation:

If \(\vec F\left( t \right)\) vector has characteristics similar to a straight line so,

\(\rm \vec F\left( t \right) \Rightarrow constant\ direction\)

so, \(\rm \vec F \times \frac{{d\vec F}}{{dt}} = 0\)

Explanation:

Curl grad ϕ = ∇ × ∇ϕ

According to property of Nable operator (∇)

∇ × ∇ϕ = 0

Where, ϕ is a scalar quantity

Let, \(\vec a = {a_1}\hat i + {a_2}\hat j + {a_3}\hat k\)

\(\begin{array}{l} \vec a \times \vec r = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k}\\ {{a_1}}&{{a_2}}&{{a_3}}\\ x&y&z \end{array}} \right|\\ \vec a \times \vec r = \hat i\left( {{a_2}z - {a_3}y} \right) - \hat j\left( {{a_1}z - {a_3}x} \right) + \hat k\left( {{a_1}y - {a_2}x} \right) \end{array}\)  

\(\nabla \times \left( {\vec a \times \vec r} \right) = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k}\\ {\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\\ {{a_2}z - {a_3}y}&{{a_3}x - {a_1}z}&{{a_1}y - {a_2}x} \end{array}} \right|\)

\(\Rightarrow \hat i\left[ {\frac{\partial }{{\partial y}}\left( {{a_1}y - {a_2}x} \right) - \frac{\partial }{{\partial z}}\left( {{a_3}x - {a_1}z} \right)} \right] - \hat j\left[ {\frac{\partial }{{\partial x}}\left( {{a_1}y - {a_2}x} \right) - \frac{\partial }{{\partial z}}\left( {{a_2}z - {a_3}y} \right)} \right] +\\\;\;\;\; \hat k\left[ {\frac{\partial }{{\partial x}}\left( {{a_3}x - {a_1}z} \right) - \frac{\partial }{{\partial y}}\left( {{a_2}z - {a_3}y} \right)} \right]\)

\(\begin{array}{l} \nabla \times \left( {\vec a \times \vec r} \right) = \hat i\left( {{a_1} + {a_1}} \right) - \hat j\left( { - {a_2} - {a_2}} \right) + \hat k\left( {{a_3} + {a_3}} \right)\\ = 2\left( {{a_1}\hat i + {a_2}\hat j + {a_3}\hat k} \right) = 2\vec a \end{array}\)  

Explanations:

The curl of the gradient of any scalar function φ is always zero. curl(grad φ) = 0

Explanation:

If a vector F is irrotational, then curl of F will be zero i.e. \(\nabla\times \vec F = 0\)

Curl of a vector

\(\vec F = \overrightarrow {{F_x}} \widehat {\dot i} + \overrightarrow {{F_y}} \widehat {\dot J} + \overrightarrow {{F_z}} \hat k\)

\(\nabla = \frac{\partial }{{\partial x}}\widehat {\dot i} + \frac{\partial }{{\partial y}}\widehat {\dot J} + \frac{\partial }{{\partial z}}\hat k\)

\(\nabla \times \vec F = Curl\left( {\vec F} \right) = \left| {\begin{array}{*{20}{c}} i&{\dot J}&k\\ {\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\\ {{F_x}}&{{F_y}}&{{F_z}} \end{array}} \right|\;\)