Question

A boat is moving due east in a region where the earth's magnetic field is 5.0×10⁻⁵ NA⁻¹ m⁻¹ due north and horizontal. The boat carries a vertical aerial 2 m long. If the speed of the boat is 1.50 ms⁻¹ , the magnitude of the induced emf in the wire of aerial is:
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Answer 1

$\huge{\boxed{\bold{Question}}}$A boat is moving due east in a region where the earth's magnetic field is 5.0×10⁻⁵ NA⁻¹ m⁻¹ due north and horizontal. The boat carries a vertical aerial 2 m long. If the speed of the boat is 1.50 ms⁻¹ , the magnitude of the induced emf in the wire of aerial is:

$\huge{\boxed{\bold{Answer}}}$

C=B_hVL

V=1.5

H=5

L=2

1.5×5×10^-5×2

$\huge{\boxed{\bold{C=0.15mv. Answer}}}$

Answer 2

$\huge\fcolorbox{black}{teal}{Solution:-}$

✏️ Given:-

★ A boat is moving due east in a region where the earth's magnetic field is 5.0×10⁻⁵ NA⁻¹ m⁻¹ due north and horizontal. The boat carries a vertical aerial of 2 m long. Also, the speed of the boat is 1.50 ms⁻¹.

✏️ To find:-

★ The magnitude of the induced emf in the wire of aerial.

✏️ Answer:-

★ The magnitude of the induced emf in the wire of aerial is : 0.15mV.

✏️ Explanation:-

★ Here,

• Flux density, B = 5 × 10^-5

• v = 1.5 m/s and

• Aerial length, l = 2m

Now,

As we all know,

Induced emf is given by:

ε = Bvl

∴ On plugging in the values, we get,

ε = 5 × 10^-5 × 1.50 × 2

⇒ε = 0.15mV

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I hope this helps! :)