A 1 m long metal rod AB completes the circuit as shown in figure. The area of circuit is perpendicular to the magnetic field of 0.10 T. If the resistance of the total circuit is 2Omega then the force needed to move the rod towards right with constant speed (v) of 1.5 m/s is ________ N.

To maintain constant speed F_ext = F_B => F_ext = i l B = ( (v B l / R) ) l B = (B^2 l^2 v / R) = ((0.1)^2 x (1)^2 x 1.5 / 2) = 7.5 x 10^-3 N

A 1 m long metal rod AB completes the circuit as shown ... | null (Physics) | DivineJEE

JEE Mains 2026 21st January Shift 1

Physics

1\text{ m}

long metal rod AB completes the circuit as shown in figure. The area of circuit is perpendicular to the magnetic field of

0.10\text{ T}

. If the resistance of the total circuit is

2\Omega

then the force needed to move the rod towards right with constant speed (v) of

1.5\text{ m/s}

is ________ N.

7.5 imes 10^{-2}

5.7 imes 10^{-3}

5.7 imes 10^{-2}

7.5 imes 10^{-3}

To maintain constant speed

F_{\text{ext}} = F_B

\Rightarrow F_{\text{ext}} = i l B

= \left( \frac{v B l}{R} \right) l B

= \frac{B^2 l^2 v}{R}

= \frac{(0.1)^2 \times (1)^2 \times 1.5}{2}

= 7.5 \times 10^{-3}\text{ N}

Select an option to instantly check whether it is correct or wrong.

Download Divine JEE App

Learn on the go with our mobile app. Access courses, live classes, and study materials anytime, anywhere.

Empowering students to achieve their IIT dreams through personalized learning and expert guidance.

contact.divinejee@gmail.com
+91 9137402346
2nd floor, C, 10, Mahant Swami Marg, nr. Pratap Stadium, Chitrakoot, Jaipur, Rajasthan 302021