Question:
A wire 2.80 m in length carries a current of 5.20 A in a region where a uniform magnetic field has a magnitude of 0.430 T. Calculate the magnitude of the magnetic force on the wire assuming the following angles between the magnetic field and the current.
(a)60 (b)90 (c)120
Answer:
(a)5.42 N (b)6.26 N (c)5.42 N
Explanation:
From the question
Length of wire (L) = 2.80 m
Current in wire (I) = 5.20 A
Magnetic field (B) = 0.430 T
Angle are different in each part.
The magnetic force is given by

So from data

Now sub parts
(a)

(b)

(c)

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Answer:
Center
Explanation:
The center is the tallest player on each team, playing near the basket. On offense, the center tries to score on close shots and rebound. But on defense, the center tries to block opponents' shots and rebound their misses.
<span>The valence electrons occur in the outer shell of the atom. </span>
Answer:
C. Mantle
Explanation:
The Mantle is a thick layer of solid and partially melted rock