First slide

Magnetic flux and induced emf

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
$v = \frac{ℓ}{t} = \frac{0.5}{2} = 0.25 {ms}^{- 1}$
Induced emf, $ε = B v ℓ = (1.0) (0.25) (0.5) = 0.125 V$
$\begin{matrix} ∴ Current in the loop, i = \frac{ε}{R} = \frac{0.125}{10} \\ = 1.25 \times 10^{- 2} A \end{matrix}$
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 x 10^-3N 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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