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 9 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 t

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.

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 t_{1}, the slope of the straight line is less steep than that of the slope from t_{1} to t_{2}.

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.

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 t

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.

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 t

As I have mentioned in post #2, the slope from t

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 t

Remember, B2 > B1 (magnitude only), so the graph has to be C.

Original post by Eimmanuel

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

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 t_{1} and from t_{1} to t_{2} are B1 and B2, respectively.

As I have mentioned in post #2, the slope from t_{1} to t_{2} is steeper than that of the slope from 0 to t_{1}, 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?

Do the same analysis from t_{1} to t_{2}. The induced emf would be –B2.

Remember, B2 > B1 (magnitude only), so the graph has to be C.

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 t

As I have mentioned in post #2, the slope from t

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 t

Remember, B2 > B1 (magnitude only), so the graph has to be C.

That makes sense, Eimmanuel. Thanks a lot.

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