A 0.5 m long metal rod PQ completes the circuit as shown in the figure. The area of the circuit is perpendicular to the magnetic field of flux density 0.15 T . If the resistance of the total circuit is 3 Ω , calculate the force needed to move the rod in the direction as indicated with a constant speed of 2 textm s−1.

Question

A 0.5 m long metal rod PQ completes the circuit as shown in the figure. The area of the circuit is perpendicular to the magnetic field of flux density 0.15 T . If the resistance of the total circuit is 3 Ω , calculate the force needed to move the rod in the direction as indicated with a constant speed of 2 textm s−1. <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-q8m2ryly7tnc.png" /> (A) 5.78mN (B) 3.75mN (C) 4.95mN (D) 11.23mN

Tardigrade Admin · Accepted Answer

F = (B2 l2 v/R) = ((. 0.15 .)2 (. 0.5 .)2 × 2/3) = 0.00375 N = 3.75 textmN

$5.78 m N$

$3.75 m N$

$4.95 m N$

$11.23 m N$

$F = \frac{B ^{2} l ^{2} v}{R} = \frac{( 0.15 ) ^{2} ( 0.5 ) ^{2} \times 2}{3}$
$= 0.00375 N$
$= 3.75 mN$

5.78mN

3.75mN

4.95mN

11.23mN

F=RB2l2v​=3(0.15)2(0.5)2×2​ =0.00375N =3.75mN

$5.78 m N$

$3.75 m N$

$4.95 m N$

$11.23 m N$

$F = \frac{B ^{2} l ^{2} v}{R} = \frac{( 0.15 ) ^{2} ( 0.5 ) ^{2} \times 2}{3}$
$= 0.00375 N$
$= 3.75 mN$