Electromagnetically launched rockets? How shocking!

In general, we have $F=I(L\times B)$ . Since the direction of current and magnetic field direction are perpendicular, we have $F=ILB=(100A)(1m)(1T)=100N$ . From Newton's second law: $F=ma$ we have $a=F/m=100N/1kg=100m/s^2$ . From laws of motion, $v=u+at=0+(100m/s^2)10s=1000~m/s$ .

the main principle involved in this problem is -" when a current carrying conductor is placed in a magentic field ,a force is experienced in the conductor"this force is called lorentz force.This principle is also used in motors and other electromagentic machines.....

lorentz force is calculated by the equation F=BILsin(theta)

B is magnetic field I is the current through the conductor L is the length of the conductor theta is the angle between length vector of the conductor and magnetic field

in this question we have I=100A L=1m B=1T theta=90 degrees

therefore force =1 100 1*sin(90) = 100N

given the mass of the conductor as 1kg accelaration experienced = F/m= 100 m/s^2

final velocity after 10 sec can then be calculated by

v=u+at u=0 therefore v=at =100*10 =1000 m/s

The total force on the bar is $\int (I \times B) \, \mathrm{d}l=(100\mathrm{A})(1\mathrm{T})(1\mathrm{m})=100\mathrm{N}$ . Thus it has a constant acceleration of $100\mathrm{N}/1\mathrm{kg}=100\mathrm{m}\,\mathrm{s}^{-2}$ . After 10 seconds, its speed would be $100\mathrm{m}\,\mathrm{s}^{-2} \times 10\mathrm{s}=1000\mathrm{m}\,\mathrm{s}^{-1}$ .

The magnitude of force on the conductors is: dF = i dl x B.

Since the flow of current in each conductor is perpendicular to the magnetic field, the \sin \theta of the cross product is replaced by unity.

Thus, dF = i dl B

However, since the conductors are all linear, the integral of the above equation yields F = i l B, for each conductor. Applying the right hand screw rule, we see that the forces on the "rails" cancel each other. The conducting element opposite the projectile must be fixed (there can be no translational motion otherwise). Thus the acceleration on the projectile is given by,

a = F/m = i l B/m = 100 A * 1 m * 1 T/ 1 kg = 100 m s^{-2}

Accelerating from rest (u=0), at this rate for 10 seconds would give a final velocity of 100 m s^{-2} * 10 s = 1000 m s^{-1}.

Force on the bar is $F=BIl$ . Also , $F=ma$ and $a=\frac{\mathrm dv}{\mathrm dt}$ . Hence , $m\frac{\mathrm dv}{\mathrm dt}=BIl \Rightarrow \mathrm dv = \frac{BIl}{m}\mathrm dt$ . Integrating with proper limits , $\int_0^v \mathrm dv = \int_0^t \frac{BIl}{m}\mathrm dt$

$\Rightarrow v=\frac{BIlt}{m}$ . Substituting the respective values , we have

$v=\frac{(1T)(100A)(1m)(10s)}{1kg}=\boxed{1000}$ m/s .

The force of a magnetic field on a current carrying wire is given by $F = BIl$ . Using the right hand rule, we calculate that the magnetic force pushes the rod to the right at all times. Now we will use Newton's Second Law to find the acceleration and the final speed of the rod.

$F = BIl = ma$

$a = \frac{BIl}{m}$

Now, kinematics:

$v = v_{0} + at$

$v = at = \frac{BIlt}{m} =\frac{1*100*1*10}{1}=1000m/s$

Let $F$ be the force acting on the rail gun, $L$ be the length of the rail-gun, $I$ be the steady current, and $B$ be the magnetic field. Then, we obtain: $\vec{F}=I (\vec{L} \times \vec{B} )$ However, since $\vec{L} \perp \vec{B}$ , the equation reduces to: $F= ILB$ The acceleration of the rail gun will then be given by: $a= \frac{F}{m}= \frac{ILB}{m}$ where $m$ is the mass of the rail-gun.

After time $t$ , the velocity of the rail gun will then be: $v_t= at = \frac{ILB}{m} t$ Plugging in the values, we obtain $v_{10 \ s}= \boxed{1000 \ m/s}$

Magnetic force on current: F = BILsinθ , F = force, B = magnetic flux in Teslas, I is the current in Ampres, L is the length of rod in metres and \theta is the angle the current makes with the magnetic field. In this case, because the rod is already perpendicular to the current, sinθ = 1

From the details given in the question, we simply sub in B = 1T, I = 100A, L = 1m to get force acting on the rod to be F = 1(100)(1) = 100N. Acceleration of the rod = F/m = 100/1 = 100 ms^{-1}.

By kinematic eqn v = u+at, we find that v = 0+(100)(10) = 1000s

We have that $1T=1 \frac{N}{A*m}$ and that $N= kg*m/s^2$ .

So $1 = \frac{1*accelleration}{100*1}$ , so accelleration = 100 m/s^2.

Final speed after $10$ seconds is therefore $10*100=1000$ m/s.

This problem deals with the concept of motional EMF. This is a typical example of how current in a loop in conjunction with a magnetic field oriented orthogonal to the plane of the loop can induce a force. The magnitude of the force on the bar is $F=Il\times B$ , where $F$ is the resultant force vector , $I$ is the current vector , and $l$ is the length of the sliding bar. Since we only really care about the magnitude of the force (if you want, you can confirm that the bar moves right according to the right hand rule), we take the magnitude of the cross product: $F=BIl\sin90^{o}=BIl$ . Thus, $F=1 T\times100 A\times1 m=100 N$ . Now, we can find the acceleration using Newton's second law: $F=ma → a=\frac{F}{m}=\frac{100 N}{1 kg}=100 \frac{m}{s^{2}}$ . Finally, we use kinematics to find the velocity after 10 seconds: $v=v_0+at → v=0+(100 \frac{m}{s^{2}})(10 s)=\fbox{1000 m/s}$ .