Recent content by likthem

Never mind, I got it. It wasn't asking for what was the magnitude of the change in magnetic field but the rate of change. Since the induced current from the field will be opposite to the that of that in the loop, the answer should be negative. So I did it correctly, just misread what the problem...

Homework Statement A square loop consists of a single turn with a resistance of 4.00 Ω. The loop has an area of 600 cm^2, and has a uniform magnetic field passing through it that is directed out of the page. The loop contains a 12-volt battery, connected as shown in the figure below...

You are given that the side of each region is 90cm (you may have a different value). In this diagram you see the regions are divided up. You can then use that info to determine what one the length of one side of one box is. Looking at the path of the particle you can see that its part of a...

Never mind, I solved my own problem. I wasn't taking into account the fact that I needed to take only a portion of the period such a quarter or half of it depending on how it traveled in a particular field. Number becomes smaller.

To solve this, you simple need to rearrange the equation that you used to get field magnitude (B) for part d, and have it solve for the particle charge (q). You should already have all the info you need to solve for it minus the field magnitude, which you can just get if solve for it using...

I got the first part of the question correct. Just make sure all of your units are correct (decimal points, powers). Unfortunately, I'm having problems with the second part of the problem. I have a similar set-up. To calculate the time that the particle spends from entering field A to...

Never mind. I figured it out. I didn't get the right answer because I forgot to multiply the one of the components of the normal force by the coefficient of static friction. Now I got the right answer. Thank you for the help!

So would I be right if I do this Ff - (cos20 F + sin 20 mg) = 0 This would be the force of friction that acts on the box, with the x component of the force acting on the box as well as the component of gravity on the slope. All this added up would have to equal to 0 to show that the box...

I have a similar problem. What I did was that to find the new force of friction I first found the normal force algebraically. Fn= (cos θ)(mg) - (sin angle) horizontal force Then I did this to find the force of friction Ff = (coefficient of static friction) [(cos θ)(mg) - (sin angle)...