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Hi, I don't understand why the answer is C, i know it's either C or D, but I don't the differentiating reason behind the y-intercept.
(edited 10 months ago)
Original post by Sadman123
Hi, I don't understand why the answer is C, i know it's either C or D, but I don't the differentiating reason behind the y-intercept.




The gradient of the given graph of B vs t differentiates C and D.
Ignore the sign of the gradient in the B vs t graph first, from 0 to t1, the slope of the straight line is less steep than that of the slope from t1 to t2.
Next include the sign of the gradient and Lenz’s law, you should be able to deduce C is the answer. If you can't, let me know and I would explain in more detail.
Reply 2
The magnitude of emf in the lower graphs should be proportional to the magnitude of the gradient of the upper graph.


Upper graph has low gradient followed by high gradient.
Reply 3
Original post by Eimmanuel



The gradient of the given graph of B vs t differentiates C and D.
Ignore the sign of the gradient in the B vs t graph first, from 0 to t1, the slope of the straight line is less steep than that of the slope from t1 to t2.
Next include the sign of the gradient and Lenz’s law, you should be able to deduce C is the answer. If you can't, let me know and I would explain in more detail.


Hm, thanks.

But can you explain in more detail, I know it's C or D, due to the EMF being the derivative of the the magnetic flux, as well as the theory of it.

So, you're saying from 0 to t_1, since the gradient (dy/dx) is less steep than t_1 to t_2. and I got nothing to go on from that.
Original post by Sadman123
Hm, thanks.

But can you explain in more detail, I know it's C or D, due to the EMF being the derivative of the the magnetic flux, as well as the theory of it.

So, you're saying from 0 to t_1, since the gradient (dy/dx) is less steep than t_1 to t_2. and I got nothing to go on from that.


Do you know what is the rate of change of magnetic flux linkage in this question?

Spoiler



The rate of change of the magnetic field is the gradient of the B vs t graph.

Let's say the magnitude of the gradients from 0 to t1 and from t1 to t2 are B1 and B2, respectively.

As I have mentioned in post #2, the slope from t1 to t2 is steeper than that of the slope from 0 to t1, so B2 is greater than B1 in magnitude.

When we consider the sign of the gradients and Lenz’s law, the induced emf from 0 to t1 would be positive. Why?

Spoiler



Do the same analysis from t1 to t2. The induced emf would be –B2.
Remember, B2 > B1 (magnitude only), so the graph has to be C.
Reply 5
Original post by Eimmanuel
Do you know what is the rate of change of magnetic flux linkage in this question?

Spoiler



The rate of change of the magnetic field is the gradient of the B vs t graph.

Let's say the magnitude of the gradients from 0 to t1 and from t1 to t2 are B1 and B2, respectively.

As I have mentioned in post #2, the slope from t1 to t2 is steeper than that of the slope from 0 to t1, so B2 is greater than B1 in magnitude.

When we consider the sign of the gradients and Lenz’s law, the induced emf from 0 to t1 would be positive. Why?

Spoiler



Do the same analysis from t1 to t2. The induced emf would be –B2.
Remember, B2 > B1 (magnitude only), so the graph has to be C.


That makes sense, Eimmanuel. Thanks a lot.

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