# Physics Exam Thoughts (1 Viewer)

#### trea99

##### Member
That's what I did. If you didn't assume it was a circle, how would you calculate area?

• pikachu975

#### Sp3ctre

##### Active Member
Was it wrong to for the flux density to use the max flux and divide it by the area of the circle loop (It didn't say the loop was circular nor that the given length was the diameter, so idk if what i did was right)
Pretty sure you had to assume that was true

• pikachu975

##### Member
I have no clue by my friend was confident to say that you had to find the "area under the curve" from the flux vs time graph - which idk if that made sense

#### Commando007

##### Member
lmfao i did pi(r) ^2 xD

• pikachu975

#### trea99

##### Member
Magnetic field strength is flux per unit area. The maximum flux on the flux vs time graph was 0.6Wb I think so the maximum magnetic field strength over the 12 seconds would be 0.6/Area of the loop. Then for part b), the rate of change of flux (flux vs time) is the gradient of the curve which is voltage and then you use the right hand grip rule or something else to find direction. I don't think you had to find the area under the flux vs time curve for anything and you really couldn't do so either because part of it was curved if I remember correctly.

• pikachu975

##### Member
Magnetic field strength is flux per unit area. The maximum flux on the flux vs time graph was 0.6Wb I think so the maximum magnetic field strength over the 12 seconds would be 0.6/Area of the loop. Then for part b), the rate of change of flux (flux vs time) is the gradient of the curve which is voltage and then you use the right hand grip rule or something else to find direction. I don't think you had to find the area under the flux vs time curve for anything and you really couldn't do so either because part of it was curved if I remember correctly.
Yeah i did exactly what you said. Yeah i don't think any of the graph was curved but to begin with I'm not sure if area under curve = flux density is a true statement to begin with

#### trea99

##### Member
Yeah i did exactly what you said. Yeah i don't think any of the graph was curved but to begin with I'm not sure if area under curve = flux density is a true statement to begin with
Yea area under Flux/time curve isn't magnetic field strength pretty sure we're correct.

• pikachu975

#### pikachu975

Magnetic field strength is flux per unit area. The maximum flux on the flux vs time graph was 0.6Wb I think so the maximum magnetic field strength over the 12 seconds would be 0.6/Area of the loop. Then for part b), the rate of change of flux (flux vs time) is the gradient of the curve which is voltage and then you use the right hand grip rule or something else to find direction. I don't think you had to find the area under the flux vs time curve for anything and you really couldn't do so either because part of it was curved if I remember correctly.
Yep finding area under the curve would mean finding the integral of the curve which is clearly wrong. 0.6/Area is correct and for part b yep you had to use faraday's law and it was 0.3 Volts and P was positive using right hand group rule

##### New Member
did anyone else get 10.9 metres for the range of the projectile?

#### trea99

##### Member
Anyone else think aligning will be better than previous years? MC was definitely harder and core had some pretty weird ones like the blackbody curve.

• cloud edwards

#### trea99

##### Member
did anyone else get 10.9 metres for the range of the projectile?
I got from memory around 9m

#### maria1999

##### New Member
I got from memory around 9m
i got 8.8. is that about right??

Sent from my iPhone using Tapatalk

##### New Member
I found time to be 2.18, then Delta X = ux T

which is 2.18 x 5 = 10.9 metres

##### Member
what did you guys get for question 17 mc:

This is the answer from solutions from competitor website lo:
"17 - B , force on magnet is downwards by the right hand slap rule, so the scale reading will be higher (I'm 90% sure). Since the unit is kilograms, it is measuring mass, so we compensate by dividing the force by 9.8."

I did this but my friends said i was wrong and it was "minus" instead of plus

• Commando007

#### trea99

##### Member
Ok, so we're given an initial velocity of 10ms^-1 and an angle of 60 to the horizontal
So initial horizontal component of velocity is 5ms^-1 (ie 10cos60) and initial vertical is 10sin60 (around 8.66)
At max height, vertical velocity is 0,
so 0 = 10sin60 - 9.8t
t = 10sin60/9.8
This is half the total time of flight
Horizontal displacement = horizontal velocity x time = 5 x 20sin60/9.8 which is around 8.8 so yea maria same answer

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• Qw0pper0

#### camstu

##### New Member
I found time to be 2.18, then Delta X = ux T

which is 2.18 x 5 = 10.9 metres
was Ux 10 or am I retarded?

#### trea99

##### Member
what did you guys get for question 17 mc:

This is the answer from solutions from competitor website lo:
"17 - B , force on magnet is downwards by the right hand slap rule, so the scale reading will be higher (I'm 90% sure). Since the unit is kilograms, it is measuring mass, so we compensate by dividing the force by 9.8."

I did this but my friends said i was wrong and it was "minus" instead of plus
I agree with you and the solution, the force on the magnet is down so there's an additional force of BIl on top of the weight force and yea we divide by 9.8 for mass

#### camstu

##### New Member
what did you guys get for question 17 mc:

This is the answer from solutions from competitor website lo:
"17 - B , force on magnet is downwards by the right hand slap rule, so the scale reading will be higher (I'm 90% sure). Since the unit is kilograms, it is measuring mass, so we compensate by dividing the force by 9.8."

I did this but my friends said i was wrong and it was "minus" instead of plus
Where did you find solutions?

##### New Member
Overall velocity was 10, so Ux was 10Cos60 which was about 8.66

• pikachu975