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- A straight wire, 0.20 m long, moves at a constant speed of 7.0 m/s
perpendicular to a magnetic field of strength 8.0×10-2 T.

A. What EMF is induced in the wire?
B. What is the direction of the induced current?
C. Determine the polarity of point A and point L
D. The wire is part of a circuit that has a resistance of 0.50 Ω. What is the
current through the wire?

A Straight Wire 020 M Long Moves At A Constant Speed Of 70 Ms Perpendicular To A Magnetic Field Of Strength 80102 T A What EMF Is Induced In The Wire B What Is class=

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Onyebuchinnajiid

(a) The emf induced in the wire is 0.112 V.

(b) The direction of the current will into the page.

(c) The polarity of A will be positive and L will be negative.

(d)  The current flowing through the wire is 0.224 A.

EMF induced in the wire

The emf induced in the wire is calculated as follows;

EMF = BLv

EMF = (8 x 10⁻²) x 0.2 x 7

EMF = 0.112 V

Direction of the current

Since, the magnetic field is out of the page, the current will follow counter clockwise and will point into the page.

Polarity of point A and point L

The current is flowing counter currently, hence the polarity of A will be positive and L will be negative.

Current through the wire

V = IR

I = V/R

I = (0.112)/0.5

I = 0.224 A

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