Answer
Verified
108.9k+ views
Hint Here we will use the concept of motional emf. when an electric conductor is moved in presence of magnetic field, an emf is induced in it (according to Lenz law), that emf is called motional emf.
It is given by expression:
$\text{e}=\left( \overrightarrow{\text{v}}\times \overrightarrow{\text{B}} \right)\times \overrightarrow{\text{l}}$
Where, v is the velocity with which it is moved.
$\overrightarrow{\text{B}}$ is magnetic field
$\text{ }\overrightarrow{\text{l}}$is length of the conductor
Complete step by step solution
Mentioned emf is given by expression
$\text{e}=\left( \overrightarrow{\text{v}}\times \overrightarrow{\text{B}} \right)\times \overrightarrow{\text{l}}$…. (1)
In this situation, if we apply the right hand thumb rule to calculate the direction of the magnetic field. It comes out to be directed into the paper (as shown in figure).
Therefore, here the magnetic field, velocity and length of the conductor are perpendicular to each other.
$\therefore $ 1 becomes,
$\text{e}=\left( \text{vB sin 90}{}^\circ \right)\times \text{l}$ (v and B are perpendicular to each other)
$\begin{align}
& =\text{vBl cos 0}{}^\circ \\
& \text{=vBl} \\
\end{align}$ [Resultant of cross product of v and B is parallel to l]
For arms CD and AB of the square frame, l is parallel to v. as (i) is also called a scalar triple product.
And if two quantities are in parallel, then it turns out to be zero.
$\therefore $No emotional emf is there.
Also,
Magnetic field reached at AD is more than magnetic field at arm BC
$\therefore $${{\text{B}}_{\text{AD}}}>{{\text{B}}_{\text{BC}}}$
Using Lenz law, one can find that, direction of current in the coil is in clockwise direction. (As magnetic field is directed inward)
Emf induced in arm AD,${{\text{e}}_{1}}={{\text{B}}_{\text{AD}}}\text{vl}$
Emf induces in arm BC,${{\text{e}}_{2}}={{\text{B}}_{\text{BC}}}\text{vl}$
Resultant emf,${{\text{e}}_{\text{net}}}=\text{vl}\left( {{\text{B}}_{1}}-{{\text{B}}_{2}} \right)$…. (2)
$\begin{align}
& {{\text{B}}_{1}}=\frac{{{\text{ }\!\!\mu\!\!\text{ }}_{0}}\text{I}}{2\text{ }\!\!\pi\!\!\text{ }{{\text{r}}_{\text{1}}}}=\frac{{{\text{ }\!\!\mu\!\!\text{ }}_{0}}\times 1}{2\text{ }\!\!\pi\!\!\text{ }\times \text{1}{{\text{0}}^{-1}}} \\
& {{\text{B}}_{2}}=\frac{{{\text{ }\!\!\mu\!\!\text{ }}_{0}}\text{I}}{2\text{ }\!\!\pi\!\!\text{ }{{\text{r}}_{2}}}=\frac{{{\text{ }\!\!\mu\!\!\text{ }}_{0}}\times 1}{2\text{ }\!\!\pi\!\!\text{ }\times \text{1}{{\text{0}}^{-1}}\times 2} \\
\end{align}$
(Where${{\text{r}}_{1}}\And {{\text{r}}_{2}}$ are distance of arm AD and BC from current carrying conductor respectively]
Substituting all values in equation (2), we get
\[\begin{align}
& {{\text{e}}_{\text{net}}}=10\times 10\times {{10}^{-2}}\left[ \frac{{{\text{ }\!\!\mu\!\!\text{ }}_{0}}}{2\text{ }\!\!\pi\!\!\text{ }\times \text{1}{{\text{0}}^{-1}}}-\frac{{{\text{ }\!\!\mu\!\!\text{ }}_{0}}}{2\text{ }\!\!\pi\!\!\text{ }\times \text{1}{{\text{0}}^{-1}}\times 2} \right] \\
& \text{ }=\frac{{{\text{ }\!\!\mu\!\!\text{ }}_{0}}\times 10}{2\text{ }\!\!\pi\!\!\text{ }}\left[ 1-\frac{1}{2} \right] \\
& \text{ }=\frac{{{\text{ }\!\!\mu\!\!\text{ }}_{\text{0}}}}{\text{4 }\!\!\pi\!\!\text{ }}\times 10 \\
\end{align}\]
\[\begin{align}
& \text{ }={{10}^{-7}}\times 10 \\
& \text{ }={{10}^{-6}} \\
& \text{ }=1\text{ }\!\!\mu\!\!\text{ V} \\
\end{align}\]
Therefore, the option (B) is correct.
Note Here we have used the formula of magnetic field due to a straight current carrying wire,
\[\text{B}=\frac{{{\text{ }\!\!\mu\!\!\text{ }}_{0}}}{4\text{ }\!\!\pi\!\!\text{ }}=\frac{2\text{I}}{\text{r}}\] Where, r is distance of observation point from conductor and I, is current moving in the conductor.
Remember following points about Lenz law:
According to Lenz’s law: The polarity of the induced emf is such that it tends to produce induced current in such a direction that it opposes the change in magnetic flux that produced it.
The ($-$) live sign in given equation e$=\left( - \right)\left( \text{d}\Phi \text{/dt} \right)$ tells about the direction
According to Lenz’s law the direction of the induced current will be such that it opposes the change in the magnetic flux.
It is given by expression:
$\text{e}=\left( \overrightarrow{\text{v}}\times \overrightarrow{\text{B}} \right)\times \overrightarrow{\text{l}}$
Where, v is the velocity with which it is moved.
$\overrightarrow{\text{B}}$ is magnetic field
$\text{ }\overrightarrow{\text{l}}$is length of the conductor
Complete step by step solution
Mentioned emf is given by expression
$\text{e}=\left( \overrightarrow{\text{v}}\times \overrightarrow{\text{B}} \right)\times \overrightarrow{\text{l}}$…. (1)
In this situation, if we apply the right hand thumb rule to calculate the direction of the magnetic field. It comes out to be directed into the paper (as shown in figure).
Therefore, here the magnetic field, velocity and length of the conductor are perpendicular to each other.
$\therefore $ 1 becomes,
$\text{e}=\left( \text{vB sin 90}{}^\circ \right)\times \text{l}$ (v and B are perpendicular to each other)
$\begin{align}
& =\text{vBl cos 0}{}^\circ \\
& \text{=vBl} \\
\end{align}$ [Resultant of cross product of v and B is parallel to l]
For arms CD and AB of the square frame, l is parallel to v. as (i) is also called a scalar triple product.
And if two quantities are in parallel, then it turns out to be zero.
$\therefore $No emotional emf is there.
Also,
Magnetic field reached at AD is more than magnetic field at arm BC
$\therefore $${{\text{B}}_{\text{AD}}}>{{\text{B}}_{\text{BC}}}$
Using Lenz law, one can find that, direction of current in the coil is in clockwise direction. (As magnetic field is directed inward)
Emf induced in arm AD,${{\text{e}}_{1}}={{\text{B}}_{\text{AD}}}\text{vl}$
Emf induces in arm BC,${{\text{e}}_{2}}={{\text{B}}_{\text{BC}}}\text{vl}$
Resultant emf,${{\text{e}}_{\text{net}}}=\text{vl}\left( {{\text{B}}_{1}}-{{\text{B}}_{2}} \right)$…. (2)
$\begin{align}
& {{\text{B}}_{1}}=\frac{{{\text{ }\!\!\mu\!\!\text{ }}_{0}}\text{I}}{2\text{ }\!\!\pi\!\!\text{ }{{\text{r}}_{\text{1}}}}=\frac{{{\text{ }\!\!\mu\!\!\text{ }}_{0}}\times 1}{2\text{ }\!\!\pi\!\!\text{ }\times \text{1}{{\text{0}}^{-1}}} \\
& {{\text{B}}_{2}}=\frac{{{\text{ }\!\!\mu\!\!\text{ }}_{0}}\text{I}}{2\text{ }\!\!\pi\!\!\text{ }{{\text{r}}_{2}}}=\frac{{{\text{ }\!\!\mu\!\!\text{ }}_{0}}\times 1}{2\text{ }\!\!\pi\!\!\text{ }\times \text{1}{{\text{0}}^{-1}}\times 2} \\
\end{align}$
(Where${{\text{r}}_{1}}\And {{\text{r}}_{2}}$ are distance of arm AD and BC from current carrying conductor respectively]
Substituting all values in equation (2), we get
\[\begin{align}
& {{\text{e}}_{\text{net}}}=10\times 10\times {{10}^{-2}}\left[ \frac{{{\text{ }\!\!\mu\!\!\text{ }}_{0}}}{2\text{ }\!\!\pi\!\!\text{ }\times \text{1}{{\text{0}}^{-1}}}-\frac{{{\text{ }\!\!\mu\!\!\text{ }}_{0}}}{2\text{ }\!\!\pi\!\!\text{ }\times \text{1}{{\text{0}}^{-1}}\times 2} \right] \\
& \text{ }=\frac{{{\text{ }\!\!\mu\!\!\text{ }}_{0}}\times 10}{2\text{ }\!\!\pi\!\!\text{ }}\left[ 1-\frac{1}{2} \right] \\
& \text{ }=\frac{{{\text{ }\!\!\mu\!\!\text{ }}_{\text{0}}}}{\text{4 }\!\!\pi\!\!\text{ }}\times 10 \\
\end{align}\]
\[\begin{align}
& \text{ }={{10}^{-7}}\times 10 \\
& \text{ }={{10}^{-6}} \\
& \text{ }=1\text{ }\!\!\mu\!\!\text{ V} \\
\end{align}\]
Therefore, the option (B) is correct.
Note Here we have used the formula of magnetic field due to a straight current carrying wire,
\[\text{B}=\frac{{{\text{ }\!\!\mu\!\!\text{ }}_{0}}}{4\text{ }\!\!\pi\!\!\text{ }}=\frac{2\text{I}}{\text{r}}\] Where, r is distance of observation point from conductor and I, is current moving in the conductor.
Remember following points about Lenz law:
According to Lenz’s law: The polarity of the induced emf is such that it tends to produce induced current in such a direction that it opposes the change in magnetic flux that produced it.
The ($-$) live sign in given equation e$=\left( - \right)\left( \text{d}\Phi \text{/dt} \right)$ tells about the direction
According to Lenz’s law the direction of the induced current will be such that it opposes the change in the magnetic flux.
Recently Updated Pages
If x2 hx 21 0x2 3hx + 35 0h 0 has a common root then class 10 maths JEE_Main
The radius of a sector is 12 cm and the angle is 120circ class 10 maths JEE_Main
For what value of x function fleft x right x4 4x3 + class 10 maths JEE_Main
What is the area under the curve yx+x1 betweenx0 and class 10 maths JEE_Main
The volume of a sphere is dfrac43pi r3 cubic units class 10 maths JEE_Main
Which of the following is a good conductor of electricity class 10 chemistry JEE_Main