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PARAGRAPH 2
In a thin rectangular metallic strip a constant current I flows along the positive x-direction, as shown in the figure. The length, width and thickness of the strip are
, w and d, respectively. A uniform magnetic field  is applied on the strip along the positive y-direction. Due to this, the charge carriers experience a net deflection along the zdirection. This results in accumulation of charge carriers on the surface PQRS and appearance of equal and opposite charges on the face opposite to PQRS. A potential difference along the z-direction is thus developed. Charge accumulation continues until the magnetic force is balanced by the electric force. The current is assumed to be uniformly distributed on the cross section of the strip and carried by electrons.
Q.
Consider two different metallic strips (1 and 2) of same dimensions (lengths , width w and thickness d)
with carrier densities n1 and n2, respectively. Strip 1 is placed in magnetic field B1 and strip 2 is placed in
magnetic field B2, both along positive y-directions. Then V1 and V2 are the potential differences developed
between K and M in strips 1 and 2, respectively. Assuming that the current I is the same for both the strips,
the correct option(s) is(are)
  • a)
    If B1 = B2 and n1 = 2n2, then V2 = 2V1
  • b)
    If B1 = B2 and n1 = 2n2, then V2 = V1
  • c)
    If B1 = 2B2 and n1 = n2, then V2 = 0.5V1
  • d)
    If B1 = 2B2 and n1 = n2, then V2 = V1
Correct answer is option 'A,C'. Can you explain this answer?
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PARAGRAPH 2In a thin rectangular metallic strip a constant current I f...
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PARAGRAPH 2In a thin rectangular metallic strip a constant current I flows along the positive x-direction, as shown in the figure. The length, width and thickness of the strip are , w and d, respectively. A uniform magnetic field is applied on the strip along the positive y-direction. Due to this, the charge carriers experience a net deflection along the zdirection. This results in accumulation of charge carriers on the surface PQRS and appearance of equal and opposite charges on the face opposite to PQRS. A potential difference along the z-direction is thus developed. Charge accumulation continues until the magnetic force is balanced by the electric force. The current is assumed to be uniformly distributed on the cross section of the strip and carried by electrons.Q.Consider two different metallic strips (1 and 2) of same dimensions (lengths , width w and thickness d)with carrier densities n1 and n2, respectively. Strip 1 is placed in magnetic field B1 and strip 2 is placed inmagnetic field B2, both along positive y-directions. Then V1 and V2 are the potential differences developedbetween K and M in strips 1 and 2, respectively. Assuming that the current I is the same for both the strips,the correct option(s) is(are)a)If B1 = B2 and n1 = 2n2, then V2 = 2V1b)If B1 = B2 and n1 = 2n2, then V2 = V1c)If B1 = 2B2 and n1 = n2, then V2 = 0.5V1d)If B1 = 2B2 and n1 = n2, then V2 = V1Correct answer is option 'A,C'. Can you explain this answer?
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PARAGRAPH 2In a thin rectangular metallic strip a constant current I flows along the positive x-direction, as shown in the figure. The length, width and thickness of the strip are , w and d, respectively. A uniform magnetic field is applied on the strip along the positive y-direction. Due to this, the charge carriers experience a net deflection along the zdirection. This results in accumulation of charge carriers on the surface PQRS and appearance of equal and opposite charges on the face opposite to PQRS. A potential difference along the z-direction is thus developed. Charge accumulation continues until the magnetic force is balanced by the electric force. The current is assumed to be uniformly distributed on the cross section of the strip and carried by electrons.Q.Consider two different metallic strips (1 and 2) of same dimensions (lengths , width w and thickness d)with carrier densities n1 and n2, respectively. Strip 1 is placed in magnetic field B1 and strip 2 is placed inmagnetic field B2, both along positive y-directions. Then V1 and V2 are the potential differences developedbetween K and M in strips 1 and 2, respectively. Assuming that the current I is the same for both the strips,the correct option(s) is(are)a)If B1 = B2 and n1 = 2n2, then V2 = 2V1b)If B1 = B2 and n1 = 2n2, then V2 = V1c)If B1 = 2B2 and n1 = n2, then V2 = 0.5V1d)If B1 = 2B2 and n1 = n2, then V2 = V1Correct answer is option 'A,C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about PARAGRAPH 2In a thin rectangular metallic strip a constant current I flows along the positive x-direction, as shown in the figure. The length, width and thickness of the strip are , w and d, respectively. A uniform magnetic field is applied on the strip along the positive y-direction. Due to this, the charge carriers experience a net deflection along the zdirection. This results in accumulation of charge carriers on the surface PQRS and appearance of equal and opposite charges on the face opposite to PQRS. A potential difference along the z-direction is thus developed. Charge accumulation continues until the magnetic force is balanced by the electric force. The current is assumed to be uniformly distributed on the cross section of the strip and carried by electrons.Q.Consider two different metallic strips (1 and 2) of same dimensions (lengths , width w and thickness d)with carrier densities n1 and n2, respectively. Strip 1 is placed in magnetic field B1 and strip 2 is placed inmagnetic field B2, both along positive y-directions. Then V1 and V2 are the potential differences developedbetween K and M in strips 1 and 2, respectively. Assuming that the current I is the same for both the strips,the correct option(s) is(are)a)If B1 = B2 and n1 = 2n2, then V2 = 2V1b)If B1 = B2 and n1 = 2n2, then V2 = V1c)If B1 = 2B2 and n1 = n2, then V2 = 0.5V1d)If B1 = 2B2 and n1 = n2, then V2 = V1Correct answer is option 'A,C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for PARAGRAPH 2In a thin rectangular metallic strip a constant current I flows along the positive x-direction, as shown in the figure. The length, width and thickness of the strip are , w and d, respectively. A uniform magnetic field is applied on the strip along the positive y-direction. Due to this, the charge carriers experience a net deflection along the zdirection. This results in accumulation of charge carriers on the surface PQRS and appearance of equal and opposite charges on the face opposite to PQRS. A potential difference along the z-direction is thus developed. Charge accumulation continues until the magnetic force is balanced by the electric force. The current is assumed to be uniformly distributed on the cross section of the strip and carried by electrons.Q.Consider two different metallic strips (1 and 2) of same dimensions (lengths , width w and thickness d)with carrier densities n1 and n2, respectively. Strip 1 is placed in magnetic field B1 and strip 2 is placed inmagnetic field B2, both along positive y-directions. Then V1 and V2 are the potential differences developedbetween K and M in strips 1 and 2, respectively. Assuming that the current I is the same for both the strips,the correct option(s) is(are)a)If B1 = B2 and n1 = 2n2, then V2 = 2V1b)If B1 = B2 and n1 = 2n2, then V2 = V1c)If B1 = 2B2 and n1 = n2, then V2 = 0.5V1d)If B1 = 2B2 and n1 = n2, then V2 = V1Correct answer is option 'A,C'. Can you explain this answer?.
Solutions for PARAGRAPH 2In a thin rectangular metallic strip a constant current I flows along the positive x-direction, as shown in the figure. The length, width and thickness of the strip are , w and d, respectively. A uniform magnetic field is applied on the strip along the positive y-direction. Due to this, the charge carriers experience a net deflection along the zdirection. This results in accumulation of charge carriers on the surface PQRS and appearance of equal and opposite charges on the face opposite to PQRS. A potential difference along the z-direction is thus developed. Charge accumulation continues until the magnetic force is balanced by the electric force. The current is assumed to be uniformly distributed on the cross section of the strip and carried by electrons.Q.Consider two different metallic strips (1 and 2) of same dimensions (lengths , width w and thickness d)with carrier densities n1 and n2, respectively. Strip 1 is placed in magnetic field B1 and strip 2 is placed inmagnetic field B2, both along positive y-directions. Then V1 and V2 are the potential differences developedbetween K and M in strips 1 and 2, respectively. Assuming that the current I is the same for both the strips,the correct option(s) is(are)a)If B1 = B2 and n1 = 2n2, then V2 = 2V1b)If B1 = B2 and n1 = 2n2, then V2 = V1c)If B1 = 2B2 and n1 = n2, then V2 = 0.5V1d)If B1 = 2B2 and n1 = n2, then V2 = V1Correct answer is option 'A,C'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of PARAGRAPH 2In a thin rectangular metallic strip a constant current I flows along the positive x-direction, as shown in the figure. The length, width and thickness of the strip are , w and d, respectively. A uniform magnetic field is applied on the strip along the positive y-direction. Due to this, the charge carriers experience a net deflection along the zdirection. This results in accumulation of charge carriers on the surface PQRS and appearance of equal and opposite charges on the face opposite to PQRS. A potential difference along the z-direction is thus developed. Charge accumulation continues until the magnetic force is balanced by the electric force. The current is assumed to be uniformly distributed on the cross section of the strip and carried by electrons.Q.Consider two different metallic strips (1 and 2) of same dimensions (lengths , width w and thickness d)with carrier densities n1 and n2, respectively. Strip 1 is placed in magnetic field B1 and strip 2 is placed inmagnetic field B2, both along positive y-directions. Then V1 and V2 are the potential differences developedbetween K and M in strips 1 and 2, respectively. Assuming that the current I is the same for both the strips,the correct option(s) is(are)a)If B1 = B2 and n1 = 2n2, then V2 = 2V1b)If B1 = B2 and n1 = 2n2, then V2 = V1c)If B1 = 2B2 and n1 = n2, then V2 = 0.5V1d)If B1 = 2B2 and n1 = n2, then V2 = V1Correct answer is option 'A,C'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of PARAGRAPH 2In a thin rectangular metallic strip a constant current I flows along the positive x-direction, as shown in the figure. The length, width and thickness of the strip are , w and d, respectively. A uniform magnetic field is applied on the strip along the positive y-direction. Due to this, the charge carriers experience a net deflection along the zdirection. This results in accumulation of charge carriers on the surface PQRS and appearance of equal and opposite charges on the face opposite to PQRS. A potential difference along the z-direction is thus developed. Charge accumulation continues until the magnetic force is balanced by the electric force. The current is assumed to be uniformly distributed on the cross section of the strip and carried by electrons.Q.Consider two different metallic strips (1 and 2) of same dimensions (lengths , width w and thickness d)with carrier densities n1 and n2, respectively. Strip 1 is placed in magnetic field B1 and strip 2 is placed inmagnetic field B2, both along positive y-directions. Then V1 and V2 are the potential differences developedbetween K and M in strips 1 and 2, respectively. Assuming that the current I is the same for both the strips,the correct option(s) is(are)a)If B1 = B2 and n1 = 2n2, then V2 = 2V1b)If B1 = B2 and n1 = 2n2, then V2 = V1c)If B1 = 2B2 and n1 = n2, then V2 = 0.5V1d)If B1 = 2B2 and n1 = n2, then V2 = V1Correct answer is option 'A,C'. Can you explain this answer?, a detailed solution for PARAGRAPH 2In a thin rectangular metallic strip a constant current I flows along the positive x-direction, as shown in the figure. The length, width and thickness of the strip are , w and d, respectively. A uniform magnetic field is applied on the strip along the positive y-direction. Due to this, the charge carriers experience a net deflection along the zdirection. This results in accumulation of charge carriers on the surface PQRS and appearance of equal and opposite charges on the face opposite to PQRS. A potential difference along the z-direction is thus developed. Charge accumulation continues until the magnetic force is balanced by the electric force. The current is assumed to be uniformly distributed on the cross section of the strip and carried by electrons.Q.Consider two different metallic strips (1 and 2) of same dimensions (lengths , width w and thickness d)with carrier densities n1 and n2, respectively. Strip 1 is placed in magnetic field B1 and strip 2 is placed inmagnetic field B2, both along positive y-directions. Then V1 and V2 are the potential differences developedbetween K and M in strips 1 and 2, respectively. Assuming that the current I is the same for both the strips,the correct option(s) is(are)a)If B1 = B2 and n1 = 2n2, then V2 = 2V1b)If B1 = B2 and n1 = 2n2, then V2 = V1c)If B1 = 2B2 and n1 = n2, then V2 = 0.5V1d)If B1 = 2B2 and n1 = n2, then V2 = V1Correct answer is option 'A,C'. Can you explain this answer? has been provided alongside types of PARAGRAPH 2In a thin rectangular metallic strip a constant current I flows along the positive x-direction, as shown in the figure. The length, width and thickness of the strip are , w and d, respectively. A uniform magnetic field is applied on the strip along the positive y-direction. Due to this, the charge carriers experience a net deflection along the zdirection. This results in accumulation of charge carriers on the surface PQRS and appearance of equal and opposite charges on the face opposite to PQRS. A potential difference along the z-direction is thus developed. Charge accumulation continues until the magnetic force is balanced by the electric force. The current is assumed to be uniformly distributed on the cross section of the strip and carried by electrons.Q.Consider two different metallic strips (1 and 2) of same dimensions (lengths , width w and thickness d)with carrier densities n1 and n2, respectively. Strip 1 is placed in magnetic field B1 and strip 2 is placed inmagnetic field B2, both along positive y-directions. Then V1 and V2 are the potential differences developedbetween K and M in strips 1 and 2, respectively. Assuming that the current I is the same for both the strips,the correct option(s) is(are)a)If B1 = B2 and n1 = 2n2, then V2 = 2V1b)If B1 = B2 and n1 = 2n2, then V2 = V1c)If B1 = 2B2 and n1 = n2, then V2 = 0.5V1d)If B1 = 2B2 and n1 = n2, then V2 = V1Correct answer is option 'A,C'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice PARAGRAPH 2In a thin rectangular metallic strip a constant current I flows along the positive x-direction, as shown in the figure. The length, width and thickness of the strip are , w and d, respectively. A uniform magnetic field is applied on the strip along the positive y-direction. Due to this, the charge carriers experience a net deflection along the zdirection. This results in accumulation of charge carriers on the surface PQRS and appearance of equal and opposite charges on the face opposite to PQRS. A potential difference along the z-direction is thus developed. Charge accumulation continues until the magnetic force is balanced by the electric force. The current is assumed to be uniformly distributed on the cross section of the strip and carried by electrons.Q.Consider two different metallic strips (1 and 2) of same dimensions (lengths , width w and thickness d)with carrier densities n1 and n2, respectively. Strip 1 is placed in magnetic field B1 and strip 2 is placed inmagnetic field B2, both along positive y-directions. Then V1 and V2 are the potential differences developedbetween K and M in strips 1 and 2, respectively. Assuming that the current I is the same for both the strips,the correct option(s) is(are)a)If B1 = B2 and n1 = 2n2, then V2 = 2V1b)If B1 = B2 and n1 = 2n2, then V2 = V1c)If B1 = 2B2 and n1 = n2, then V2 = 0.5V1d)If B1 = 2B2 and n1 = n2, then V2 = V1Correct answer is option 'A,C'. Can you explain this answer? tests, examples and also practice JEE tests.
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