First slide

Magnetic force

Question

A rod of mass 0.720 kg and radius 6 cm rests on two parallel rails that are $d=12 cm$ apart and length of rails is $L=49 cm$ . The rod carries a current of $i=48 A$ in the direction shown in figure. And rolls along the rails without slipping. A uniform magnetic field of magnitude 0.240 T is directed perpendicular to the rod and the rails. If it starts from rest, what is the speed of the rod as it leaves the rails.

Difficult

A

1.12 m/s
B

2.12 m/s
C

3.12 m/s
D

4.12 m/s

Solution

$U \sin g w o r k e n e r g y t h e o r e m, W_{b y M a g n e t i c f o r c e} + W_{b y f r i c t i o n} = Δ K E \Rightarrow FL= \frac{1}{2} m v^{2} + \frac{1}{2} I w^{2} [S t a t i c f r i c t i o n d o e s n o w o r k d u r i n g p u r e r o l l i n g] \Rightarrow Bid L = \frac{1}{2} m v^{2} + \frac{1}{2} (\frac{1}{2} m r^{2}) {(\frac{v}{r})}^{2} ∴ V = \sqrt{\frac{4}{3} \frac{i d B L}{m}} = \sqrt{\frac{4 \times 48 \times 0.12 \times 0.240 \times 0.49}{3 \times 0.720}} = 1.12 m/s$

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