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An electron travels anticlockwise on a square path of side 20cm at a constant speed of 5×10² m/s (i) Find the current at a point on the square if voltage across side of square is 10V. (ii) Also find resistance of one side of square. (Take charge on electron = 1.6 × 10^-19)?
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An electron travels anticlockwise on a square path of side 20cm at a c...
**Given:**
Side of square, a = 20cm = 0.2m
Speed of electron, v = 5×10² m/s
Voltage across side of square, V = 10V
Charge on electron, q = 1.6 × 10^-19 C
**To find:**
(i) Current at a point on the square
(ii) Resistance of one side of square
**Solution:**
**(i) Current at a point on the square:**
The current can be found using the formula:
I = V/R
where I is the current, V is the voltage across the side of the square and R is the resistance of one side of the square.
To find the current, we need to first find the resistance of one side of the square.
**(ii) Resistance of one side of square:**
The resistance of one side of the square can be found using the formula:
R = ρL/A
where R is the resistance, ρ is the resistivity of the material, L is the length of the side of the square and A is the area of cross-section of the wire.
As the material of the wire is not given, we assume that it is made up of copper. The resistivity of copper is 1.7 × 10^-8 Ωm.
Length of the side of the square, L = a = 0.2m
Area of cross-section of the wire, A = (π/4)d^2, where d is the diameter of the wire
As the diameter of the wire is not given, we assume that it is a thin wire, so we take d = 0.1mm = 1 × 10^-4 m
Therefore, A = (π/4)(1 × 10^-4)^2 = 7.85 × 10^-9 m^2
Now, substituting the values in the formula, we get:
R = (1.7 × 10^-8 × 0.2)/7.85 × 10^-9 = 0.43 Ω
**(i) Current at a point on the square (continued):**
Now that we have found the resistance, we can find the current using the formula:
I = V/R
Substituting the values, we get:
I = 10/0.43 = 23.26 A
Therefore, the current at a point on the square is 23.26 A.
**Conclusion:**
- The resistance of one side of the square is 0.43 Ω.
- The current at a point on the square is 23.26 A.
Side of square, a = 20cm = 0.2m
Speed of electron, v = 5×10² m/s
Voltage across side of square, V = 10V
Charge on electron, q = 1.6 × 10^-19 C
**To find:**
(i) Current at a point on the square
(ii) Resistance of one side of square
**Solution:**
**(i) Current at a point on the square:**
The current can be found using the formula:
I = V/R
where I is the current, V is the voltage across the side of the square and R is the resistance of one side of the square.
To find the current, we need to first find the resistance of one side of the square.
**(ii) Resistance of one side of square:**
The resistance of one side of the square can be found using the formula:
R = ρL/A
where R is the resistance, ρ is the resistivity of the material, L is the length of the side of the square and A is the area of cross-section of the wire.
As the material of the wire is not given, we assume that it is made up of copper. The resistivity of copper is 1.7 × 10^-8 Ωm.
Length of the side of the square, L = a = 0.2m
Area of cross-section of the wire, A = (π/4)d^2, where d is the diameter of the wire
As the diameter of the wire is not given, we assume that it is a thin wire, so we take d = 0.1mm = 1 × 10^-4 m
Therefore, A = (π/4)(1 × 10^-4)^2 = 7.85 × 10^-9 m^2
Now, substituting the values in the formula, we get:
R = (1.7 × 10^-8 × 0.2)/7.85 × 10^-9 = 0.43 Ω
**(i) Current at a point on the square (continued):**
Now that we have found the resistance, we can find the current using the formula:
I = V/R
Substituting the values, we get:
I = 10/0.43 = 23.26 A
Therefore, the current at a point on the square is 23.26 A.
**Conclusion:**
- The resistance of one side of the square is 0.43 Ω.
- The current at a point on the square is 23.26 A.
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An electron travels anticlockwise on a square path of side 20cm at a constant speed of 5×10² m/s (i) Find the current at a point on the square if voltage across side of square is 10V. (ii) Also find resistance of one side of square. (Take charge on electron = 1.6 × 10^-19)?
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An electron travels anticlockwise on a square path of side 20cm at a constant speed of 5×10² m/s (i) Find the current at a point on the square if voltage across side of square is 10V. (ii) Also find resistance of one side of square. (Take charge on electron = 1.6 × 10^-19)? for Class 10 2024 is part of Class 10 preparation. The Question and answers have been prepared according to the Class 10 exam syllabus. Information about An electron travels anticlockwise on a square path of side 20cm at a constant speed of 5×10² m/s (i) Find the current at a point on the square if voltage across side of square is 10V. (ii) Also find resistance of one side of square. (Take charge on electron = 1.6 × 10^-19)? covers all topics & solutions for Class 10 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An electron travels anticlockwise on a square path of side 20cm at a constant speed of 5×10² m/s (i) Find the current at a point on the square if voltage across side of square is 10V. (ii) Also find resistance of one side of square. (Take charge on electron = 1.6 × 10^-19)?.
An electron travels anticlockwise on a square path of side 20cm at a constant speed of 5×10² m/s (i) Find the current at a point on the square if voltage across side of square is 10V. (ii) Also find resistance of one side of square. (Take charge on electron = 1.6 × 10^-19)? for Class 10 2024 is part of Class 10 preparation. The Question and answers have been prepared according to the Class 10 exam syllabus. Information about An electron travels anticlockwise on a square path of side 20cm at a constant speed of 5×10² m/s (i) Find the current at a point on the square if voltage across side of square is 10V. (ii) Also find resistance of one side of square. (Take charge on electron = 1.6 × 10^-19)? covers all topics & solutions for Class 10 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An electron travels anticlockwise on a square path of side 20cm at a constant speed of 5×10² m/s (i) Find the current at a point on the square if voltage across side of square is 10V. (ii) Also find resistance of one side of square. (Take charge on electron = 1.6 × 10^-19)?.
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