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sasho [114]
3 years ago
11

A single loop of current is immersed in an externally applied uniform magnetic field of 3 Tesla oriented in the positive y direc

tion: begin mathsize 12px style B with rightwards arrow on top equals 3 space j with logical and on top end style T. The loop carries a 1/2 amp current, I = 0.5 A. If the lengths of the sides of the loop are 4 m and 2 m, what is the magnitude of the magnetic moment, mu, of the loop?
Physics
1 answer:
vlabodo [156]3 years ago
6 0

Answer:

mu=12Tm^2

Explanation:

the magnetic moment mu of a single loop is given by:

\mu = I A B

where I is the current, B is the magnetic field and A is the area of the loop. By replacing we obtain:

\mu=(0.5A)(4m*2m)(3T)=12Tm^2

hope this helps!!

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A square loop of side 7 cm is placed with the nearest side 2 cm from a long wire carrying a current that varies with time at a c
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Answer:

Explanation:

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di/dt = 9 A/s

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\int _{0}^{i}di =9\int _{0}^{t}dt

i = 9t

(a) The magnetic field due to the current carrying wire at a distance r is given by

B = \frac{\mu_{0}i}{2\pi r}

B = \frac{\mu_{0}\times 9t}{2\pi r}

(b)

Magnetic flux,

\phi=\int B\times a dr

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