
The electric field intensity at all points in space is given by $\vec E = \sqrt 3 \hat i - \hat j$ volts/ampere. A square frame LMNO of side 1 meter is shown in figure. The point N lies in the x-y plane. The initial angle between line ON and x-axis is $\theta = 60^\circ $. What is the magnitude of electric flux through the area enclosed in square frame LMNO?

A) 0 volt meter
B) 1 volt meter
C) 2 volt meter
D) 4 volt meter
Answer
131.4k+ views
Hint: The magnitude of the electric flux through the square frame is the dot product of the electric field intensity with the area of the frame.
Complete step by step solution:
The electric field intensity is given to us. The electric field intensity does not vary with position or time, thus it is a constant vector. The area of the square frame is 1 ${m^2}$. But the frame is at an angle of 60 degrees with the x-axis. So the area vector, denoted by $\vec S$, becomes,
$\vec S = 1\sin \left( {60^\circ } \right)\hat i + 1\cos \left( {60^\circ } \right)\left( { - \hat j} \right) = \dfrac{{\sqrt 3 }}{2}\hat i - \dfrac{1}{2}\hat j$
Now, we can find out the electric flux passing through the area enclosed by the square frame by taking the dot product of electric field intensity and area vector. On taking dot product, we obtain,
$\phi = \vec E \cdot \vec S$ …equation (1)
We now substitute the values of the vectors of area and electric field intensity in equation (1). We get,
$\phi = \left( {\sqrt 3 \hat i - \hat j} \right) \cdot \left( {\dfrac{{\sqrt 3 }}{2}\hat i - \dfrac{1}{2}\hat j} \right) = \dfrac{3}{2} + \dfrac{1}{2} = 2$ volt meter
Hence, the electric flux passing through the square frame is 2 volt meter.
Therefore, the correct answer of the question is option (C).
Note: The above question can also be solved by finding out the magnitude of the electric field intensity vector and area vector and then finding out the angle between these vectors to find out the electric flux passing through the square frame. But, doing that is more prone to errors because it will be relatively more calculation intensive.
Complete step by step solution:
The electric field intensity is given to us. The electric field intensity does not vary with position or time, thus it is a constant vector. The area of the square frame is 1 ${m^2}$. But the frame is at an angle of 60 degrees with the x-axis. So the area vector, denoted by $\vec S$, becomes,
$\vec S = 1\sin \left( {60^\circ } \right)\hat i + 1\cos \left( {60^\circ } \right)\left( { - \hat j} \right) = \dfrac{{\sqrt 3 }}{2}\hat i - \dfrac{1}{2}\hat j$
Now, we can find out the electric flux passing through the area enclosed by the square frame by taking the dot product of electric field intensity and area vector. On taking dot product, we obtain,
$\phi = \vec E \cdot \vec S$ …equation (1)
We now substitute the values of the vectors of area and electric field intensity in equation (1). We get,
$\phi = \left( {\sqrt 3 \hat i - \hat j} \right) \cdot \left( {\dfrac{{\sqrt 3 }}{2}\hat i - \dfrac{1}{2}\hat j} \right) = \dfrac{3}{2} + \dfrac{1}{2} = 2$ volt meter
Hence, the electric flux passing through the square frame is 2 volt meter.
Therefore, the correct answer of the question is option (C).
Note: The above question can also be solved by finding out the magnitude of the electric field intensity vector and area vector and then finding out the angle between these vectors to find out the electric flux passing through the square frame. But, doing that is more prone to errors because it will be relatively more calculation intensive.
Recently Updated Pages
Young's Double Slit Experiment Step by Step Derivation

Difference Between Circuit Switching and Packet Switching

Difference Between Mass and Weight

JEE Main Participating Colleges 2024 - A Complete List of Top Colleges

JEE Main Maths Paper Pattern 2025 – Marking, Sections & Tips

Sign up for JEE Main 2025 Live Classes - Vedantu

Trending doubts
Degree of Dissociation and Its Formula With Solved Example for JEE

Displacement-Time Graph and Velocity-Time Graph for JEE

Clemmenson and Wolff Kishner Reductions for JEE

Sir C V Raman won the Nobel Prize in which year A 1928 class 12 physics JEE_Main

In Bohrs model of the hydrogen atom the radius of the class 12 physics JEE_Main

JEE Main 2025 Session 2 Registration Open – Apply Now! Form Link, Last Date and Fees

Other Pages
JEE Advanced 2024 Syllabus Weightage

CBSE Date Sheet 2025 Class 12 - Download Timetable PDF for FREE Now

JEE Main 2025 - Session 2 Registration Open | Exam Dates, Answer Key, PDF

CBSE Class 10 Hindi Sample Papers with Solutions 2024-25 FREE PDF

CBSE Board Exam Date Sheet Class 10 2025 (OUT): Download Exam Dates PDF

CBSE Class 10 Hindi Course-B Syllabus 2024-25 - Revised PDF Download
