Question

rectangular coil of resistance R1=12.91 ohms has N1=182.23 turns, each with length L1=6.53*10^-03 m and width w1=1.03*10^-02 m, as shown in the figure. The coil moves into a uniform magnetic field of B1=0.31 T with constant velocity v1=31.12 m/s. What is the magnitude of the total magnetic force F on the coil as it leaves the field? Give the answer in N.​

Answer 1

Answer:

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Answer 2

Given:

Resistance of coil $R=12.91 ohm$

Number of turns $N=182.23 turns$

Length of coil $L=6.53$×$10^{-3}$ m

Width of coil $w=1.03$×$10^{-2}$ m

Magnetic field $B=0.31 T$

Velocity $v=31.12m/s$

To Find:

F=?

Formula:

$F=NBIL$

Solution:

Step 1 of 2

The magnitude of magnetic force is determined by the product of number of turns, magnetic field, current and length of coil.

But as the coil is inserting vertically, so the formula will be changed as

$F=NBIw$

The current in the coil is not given in the question, so we need to use Faraday's law to determine the emf in the coil.

Faraday's law of electromagnetic induction $e=\frac{d(phi)}{dt}$

$phi=NBA\\A=Area\\phi=NBwL$

$e=\frac{NBwdL}{dt}=NBwv$

The current in the coil can be determined by dividing the emf with the resistance in the coil.

$I=\frac{e}{R}=$$\frac{NBwv}{R}$

Step 2 of 2

Thus, the formula for determining the force acting on coil leaving the field will be

$F=NBw$×$I$

$F=NBw$×$\frac{NBwv}{R}$

$F=\frac{N^{2}B^{2}w^{2}v}{R}$

Now, substitute the values given in the question for the above parameters to determine the force.

$F=(182.23)^{2}$×$(0.31)^{2}$×$(1.03)^{2}$×$(10^{-2})^{2}$×$\frac{31.12}{12.91}$

$F=33207.77$×$0.0961$×$1.0609$×$10^{-4}$×$2.41$

$F=8183.03 N$

So, the magnitude of total magnetic force on the coil as it leaves the field is 8183.03 N.

Answer:

The magnitude of total magnetic force on the coil as it leaves the field is 8183.03N.