Question

6. Magneto, the villain from X-Men, does not actually need metal to cause mayhem. Magneto is quite capable of using the phenomenon of magnetic induction in order to induce currents through his victims. a Magneto creates a massive 400T magnetic field through his adversary Bob, who is 1.8m tall and immediately begins sprinting away from Magneto at a rate of6.0 m/s. Stupidly, Bob is fleeing in a direction exactly perpendicular to Magneto's magnetic field. How much current is induced through poor Bob if his head-to-toe resistance is R 270 Q? Bob b) Using the diagram, will that current flow toward Bob's feet or toward his head? c) Is this current pushed by a magnetic force or by an electric force? (In other words, which of the two types of induction is being activated: motional or transformer?) Having easily dispensed with Bob, Magneto turns to the real threat: Amy. She freezes, realizing that the magnetic force trying to push current through her will be zero as long as her velocity is zero. Magnetism exerts no force on motionless charges! But Magneto has a trick up his sleeve., d) Suppose Amy is clasping her hands together, forming a loop of area 0.25m .IfMagneto steadily tums his 400T magnetic field offover the course ofjust 80 milliseconds, how much current will be induced in Amy's arms during that time? The total resistance of Amy's arm loop is 1000. B, decreasing

Answer 1

Answer:-

Part -a) Given Data:-

B= 400T

L =1.8 m

V = 0.6 m/s

R =270 ohm.

Direction is exactely perpendicular so $\theta$ = 90⁰

first find emf by using formula

emf $\varepsilon$ = vBL sin $\theta$

= 0.6*400*1.8*sin90

emf = 432 V

now calculate the induced current I = emf/R = 432 / 270

I = 1.6 Amp.

part-b) By using diagram we can say that the current flowing toward his head.

Part-c) This current push by a magnetic force F = I*LB sin $\theta$ .

Part-d) We have to find the induced current

Given data

B= 400 T

A =0.25 m2

T =80 milisecond = 0.08 sec

emf = d (B*A) /dt

= 400*0.25 /0.08

= 1250 V

So induced current I = emf/R

I = 1250 /100 = 12.5 Amp

part-e) relative to the diagram we can say that the current will flow in clokcwise direction.

Part-f) This current will pushed by magnetic force.