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 l, w and d, respectively. A uniform magnetic field →B is applied on the strip along the positive y-direction. Due to this, the charge carriers experience a net deflection along the z-direction. 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.
Consider two different metallic strips (1 and 2) of the same material. Their lengths are the same, widths are w1 and w2 and thicknesses are d1 and d2, respectively. Two points K and M are symmetrically located on the opposite faces parallel to the x-y plane (see figure). V1 and V2 are the potential differences between K and M in strips 1 and 2, respectively. Then, for a given current I flowing through them in a given magnetic field strength B, the correct statement(s) is(are),
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 l, w and d, respectively. A uniform magnetic field →B is applied on the strip along the positive y-direction. Due to this, the charge carriers experience a net deflection along the z-direction. 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.
Consider two different metallic strips (1 and 2) of the same material. Their lengths are the same, widths are w1 and w2 and thicknesses are d1 and d2, respectively. Two points K and M are symmetrically located on the opposite faces parallel to the x-y plane (see figure). V1 and V2 are the potential differences between K and M in strips 1 and 2, respectively. Then, for a given current I flowing through them in a given magnetic field strength B, the correct statement(s) is(are),
The correct option is D If w1=2w2 and d1=d2, then V2=V1.
The charge carriers in a metallic strip are electrons. The electrons moves in a direction opposite to the current flow with a drift velocity given by, vd=IneA=Inewd, where n is the number density of the electrons (number of electrons per unit volume), e is the electronic charge, and A=wd is the cross-sectional area of the strip. The electrons move with velocity →vd=−vd^ı in a uniform magnetic field →B=B^ȷ. Thus, magnetic force on the moving electrons is, →Fm=q→vd×→B=(−e)(−vd^ı)×(B^ȷ)=evdB^k=IBnwd^k. Note that an electron has a negative charge q=−e. From above equation, the electrons start moving towards the face PQRS leading to the negative potential at this face. This potential establishes an electric field in a direction from K to M. If V is the potential difference between these faces then electric field is →E=(V/w)^k. The electric force on the electrons is, →Fe=q→E=(−e)Vw^k=−eVw^k. The accumulation of electrons on face PQRS continues till the net force on the electrons become zero (equilibrium condition). Use equations to get the potential difference in equilibrium condition, V=IBned. From this equation, V depends on d and not on w. Thus, if d1=d2 then V2=V1 and if d1=2d2 then V2=2V1. Thus, A and D are correct answers.
The charge carriers in a metallic strip are electrons. The electrons moves in a direction opposite to the current flow with a drift velocity given by, vd=IneA=Inewd, where n is the number density of the electrons (number of electrons per unit volume), e is the electronic charge, and A=wd is the cross-sectional area of the strip. The electrons move with velocity →vd=−vd^ı in a uniform magnetic field →B=B^ȷ. Thus, magnetic force on the moving electrons is, →Fm=q→vd×→B=(−e)(−vd^ı)×(B^ȷ)=evdB^k=IBnwd^k. Note that an electron has a negative charge q=−e. From above equation, the electrons start moving towards the face PQRS leading to the negative potential at this face. This potential establishes an electric field in a direction from K to M. If V is the potential difference between these faces then electric field is →E=(V/w)^k. The electric force on the electrons is, →Fe=q→E=(−e)Vw^k=−eVw^k. The accumulation of electrons on face PQRS continues till the net force on the electrons become zero (equilibrium condition). Use equations to get the potential difference in equilibrium condition, V=IBned. From this equation, V depends on d and not on w. Thus, if d1=d2 then V2=V1 and if d1=2d2 then V2=2V1. Thus, A and D are correct answers.
