First slide

Energy consideration quantitative study.

Question

Figure shows a square loop of side 0.5 m and resistance $10 Ω .$ The magnetic field has a magnitude B = 1.0 T. The work done in pulling the loop out of the field slowly and uniformly in 2.0 s is

Moderate

A

$3.125 \times 10^{- 3} J$
B

$6.25 \times 10^{- 4} J$
C

$1.25 \times 10^{- 2} J$
D

$5.0 \times 10^{- 4} J$

Solution

Speed of the loop should be
$v = \frac{ℓ}{t} = \frac{0.5}{2} = 0.25 m / s$
Induced emf, $e = B v ℓ = (1.0) (1.0) (0.25) (0.5)$
= 0.125 V
$∴$ Current in the loop, $i = \frac{e}{R} = \frac{0.125}{10}$
$= 1.25 \times 10^{- 2} A$
The magnetic force on the left arm due to the magnetic field is
$F_{m} = i ℓ B = (1.25 \times 10^{- 2}) (0.5) (1.0)$
$= 6.25 \times 10^{- 3} N$
To pull the loop uniformly, an external force of $6.25 \times 10^{- 3} N$ towards right must be applied.
$∴ W = (6 .25 \times 10^{- 3} N) (0 .5 m) = 3 .125 \times 10^{- 3} J$

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