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olga55 [171]
3 years ago
5

We move a square loop of wire 2.3 cm on a side into a region of uniform magnetic field of 1.5 T with the plane of the loop perpe

ndicular to the direction of the field. The loop is moving with a constant velocity of 2.1 m/s into the field region starting from a point outside the field. If the loop has a resistance of 3 Ω, find the force on the loop as it just enters the magnetic field region. (Enter the magnitude.)
Physics
1 answer:
vova2212 [387]3 years ago
4 0

Answer:

Force will be F=8.33\times 10^{-4}N

Explanation:

We have given side of wire d = 2.3 cm = 0.023 m

Magnetic field B = 1.5 T

Velocity v = 2.1 m/sec

Resistance R = 3 ohm

We know that current is given by i=\frac{Bvd}{R}=\frac{1.5\times 2.1\times 0.023}{3}=0.02415A

Power is given by P=i^2R=0.02415^2\times 3=0.00174watt

We know that power is also given by P=Fv

So 0.00174=F\times 2.1

F=8.33\times 10^{-4}N

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Answer:

Explanation:

Given that

The electric fields of strengths E = 187,500 V/m and

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The diameter d is 25.05 cm which is converted to 0.2505m

The radius is (d/2)

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The given formula to find the magnetic force is F_{ma}=BqV---(i)

The given formula to find the electric force is F_{el}=qE---(ii)

The velocity of electric field and magnetic field is said to be perpendicular

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Equate equation (i) and equation (ii)

Bqv=qE\\\\v=\frac{E}{B}

v=\frac{187500}{0.125} \\\\v=15\times10^5m/s

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F_{ce}=\frac{mv^2}{r}---(iii)

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Bqr=\frac{mv^2}{r} \\\\\frac{q}{m}=\frac{v}{Br}  \\\\\frac{q}{m} =\frac{15\times 10^5}{0.125\times0.12525} \\\\=\frac{15\times10^5}{0.015656} \\\\=95808383.23\\\\=958.1\times10^5C/kg

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b)

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Then 1/ 958.1 × 10⁵ C/kg

The charge to mass ratio is very close to that of a proton, which is about 1*10^8 C/kg

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