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suter [353]
2 years ago
7

A circular loop ( radius = 0.5 m) carries a current of 3.0 A and has unit normal vector of (2i - j +2k)/3 what is x component of

the torque on this loop when it is placed in a uniform magnetic field of (2i-6j) T
Physics
1 answer:
Roman55 [17]2 years ago
7 0

Answer:

T=9.42Nm

Explanation:

From the question we are told that:

Radius r= 0.5 m

Current I= 3.0 A

Normal vector n=\frac{(2i - j +2k)}{3}

Magnetic field B= (2i-6j) T

Generally the equation for Area is mathematically given by

 A=\pi r^2

 A=3.1415 *0.5^2

 A=0.7853 m^2

Generally the equation for Torque is mathematically given by

 T=A(i'*B)

Where

 i'*B= \begin{bmatrix}2&-1&2\\2&-6&0\end{bmatrix}

 X\ component\ of\ i'*B= [(-1 * 0)-(2*-6)]

 X\ component\ of\ i'*B=12

Therefore

 T=0.7853*12

 T=9.42Nm

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3 years ago
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<u>Answer:</u> The ball is travelling with a speed of 5.5 m/s after hitting the <u>bottle.</u>

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We are given:

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