A rigid conducting rod of length L is pivoted at one end O and rotates with constant angular velocity ω in a plane. A non-uniform magnetic field B = B 0 ( 1 + L r ) k ^ exists in the region, where

Question

A rigid conducting rod of length L is pivoted at one end O and rotates with constant angular velocity ω in a plane. A non-uniform magnetic field B = B 0 ​ ( 1 + L r ​ ) k ^ exists in the region, where r is the distance from O and k ^ is perpendicular to the plane of rotation. Match the quantities in List-I with their corresponding expressions in List-II. List - I List - II (P) EMF induced across the full length of the rod (between O and the other end) (1) 6 B 0 ​ ω L 2 ​ (Q) Magnitude of the electric field induced at the midpoint of the rod ( r = L /2 ) (2) 6 5 B 0 ​ ω L 2 ​ (R) Potential difference between the pivot O and a point P on the rod at r = L /2 (3) 2 3 I B 0 ​ L ​ (S) If the rod is connected to an external resistor R e x t ​ such that a current I flows from O to L, the magnitude of the total magnetic force on the rod (4) 4 3 B 0 ​ ω L ​ (5) 2 B 0 ​ ω L 2 ​ (6) B 0 ​ ω L 2

TestHub · Accepted Answer

The motional EMF is E = ∫ 0 L ​ ( v × B ) ⋅ d l . With v = ω r θ ^ and B = B 0 ​ ( 1 + r / L ) k ^ , we get v × B = B 0 ​ ω r ( 1 + r / L ) r ^ . Integration from 0 to L yields E = 6 5 ​ B 0 ​ ω L 2 . So (P)-(2). The induced electric field due to charge separation is E = − ( v × B ) , so its magnitude is B 0 ​ ω r ( 1 + r / L ) . At r = L /2 , this is B 0 ​ ω ( L /2 ) ( 1 + 1/2 ) = 4 3 ​ B 0 ​ ω L . So (Q)-(4). Potential difference V P ​ − V O ​ = ∫ 0 L /2 ​ ( v × B ) ⋅ d r = ∫ 0 L /2 ​ B 0 ​ ω ( r + r 2 / L ) d r = B 0 ​ ω [ r 2 /2 + r 3 / ( 3 L ) ] 0 L /2 ​ = B 0 ​ ω ( L 2 /8 + L 2 /24 ) = 6 1 ​ B 0 ​ ω L 2 . So (R)-(1). The magnetic force on a current-carrying segment d l is d F = I d l × B . For d l = d r r ^ , d F = I ( d r r ^ ) × B 0 ​ ( 1 + r / L ) k ^ = − I B 0 ​ ( 1 + r / L ) d r θ ^ . Integrating from 0 to L gives total force magnitude I B 0 ​ [ r + r 2 / ( 2 L ) ] 0 L ​ = I B 0 ​ ( L + L /2 ) = 2 3 ​ I B 0 ​ L . So (S)-(3).

List - I	List - II
(P) EMF induced across the full length of the rod (between O and the other end)	(1) $\frac{B _{0} ω L ^{2}}{6}$
(Q) Magnitude of the electric field induced at the midpoint of the rod ( $r = L /2$ )	(2) $\frac{5 B _{0} ω L ^{2}}{6}$
(R) Potential difference between the pivot O and a point P on the rod at $r = L /2$	(3) $\frac{3 I B _{0} L}{2}$
(S) If the rod is connected to an external resistor $R_{e x t}$ such that a current $I$ flows from O to L, the magnitude of the total magnetic force on the rod	(4) $\frac{3 B _{0} ω L}{4}$
	(5) $\frac{B _{0} ω L ^{2}}{2}$
	(6) $B_{0} ω L^{2}$

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