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A single-turn current loop, carrying a current of 4.03 A, is in the shape of a right triangle with sides 68.1, 151, and 166 cm. The loop is in a uniform magnetic field of magnitude 99.3 mT whose direction is parallel to the current in the 166 cm side of the loop. What is the magnitude of the magnetic force on
(a) the 166 cm side,
(b) the 68.1 cm side, and
(c) the 151 cm side?
(d) What is the magnitude of the net force on the loop?

Respuesta :

Answer:

Part a)

[tex]F = 0[/tex]

Part b)

[tex]F = 0.25 N[/tex]

Part c)

[tex]F = 0.25 N[/tex]

Part d)

Net force on a closed loop in uniform magnetic field is always ZERO

[tex]F_{net} = 0[/tex]

Explanation:

As we know that force on a current carrying wire is given as

[tex]\vec F = i(\vec L \times \vec B)[/tex]

now we have

Part a)

current in side 166 cm and magnetic field is parallel

so we have

[tex]F = i(\vec L \times \vec B)[/tex]

here we know that L and B is parallel to each other so

[tex]F = 0[/tex]

Part b)

For 68.1 cm length wire we have

[tex]F = iLB sin\theta[/tex]

here we know that

[tex]cos\theta = \frac{68.1}{166}[/tex]

[tex]\theta = 65.8[/tex]

so we have

[tex]F = (4.03)(0.681)(99.3 \times 10^{-3})sin65.8[/tex]

[tex]F = 0.25 N[/tex]

Part c)

For 151 cm length wire we have

[tex]F = iLB sin\phi[/tex]

here we know that

[tex]cos\phi = \frac{151}{166}[/tex]

[tex]\theta = 24.5[/tex]

so we have

[tex]F = (4.03)(1.51)(99.3 \times 10^{-3})sin24.5[/tex]

[tex]F = 0.25 N[/tex]

Part d)

Net force on a closed loop in uniform magnetic field is always ZERO

[tex]F_{net} = 0[/tex]

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