What is the Induced Current in the Second Coil?

[cashcabwatche]

Homework Statement

The mutual inductance between two coils is 0.0100 H. The current in the first coil changes uniformly from 2.7 A to 5.0 A in 0.160 s. If the second coil has a resistance of 0.600 Ω, what is the magnitude of the induced current in the second coil?

Please show steps

Homework Equations

I=V/R
V=M(di/dt)

The Attempt at a Solution

I tried everything but I still couldn't figure out the magnitude of the induced current in the second coil

What I did first is I solved for V_2 using M(di_1/dt)
di_1=(5-2.7)=2.3 A
dt =0.16 s
M=.01 H
and I got V_2 to be around .14

then once I got V_2 I divided it by .6 and got I to be around .23A but my friend keeps on getting the answer to be .344 A

 

[berkeman]

Homework Statement

The mutual inductance between two coils is 0.0100 H. The current in the first coil changes uniformly from 2.7 A to 5.0 A in 0.160 s. If the second coil has a resistance of 0.600 Ω, what is the magnitude of the induced current in the second coil?

Please show steps

Homework Equations

I=V/R
V=M(di/dt)

The Attempt at a Solution

I tried everything but I still couldn't figure out the magnitude of the induced current in the second coil

Welcome to the PF. Please show us what you have tried so far, and we can offer some hints after seeing some of your work.

 

[rock.freak667]

How about we change V=M(dI/dt) to this

$$V = M( \frac{Change \ in \ Current}{Change \ in \ time})$$

 

[cashcabwatche]

See I tried that and still got the answer I'm getting

 

[berkeman]

See I tried that and still got the answer I'm getting

I will say this one more time. You need to show us your work before we can offer tutorial help. That's part of the PF Rules (see the link at the top of the page).

EDIT -- Oh, I see you went back and edited your original post (OP) to show some work. Pretty stealthy. It's usually better to just post the new work as a new post. Let me go back and take a look...

 

[cashcabwatche]

I already did show work!

 

[rock.freak667]

See I tried that and still got the answer I'm getting

Well your method looks correct to me.