First slide

Induced Electric Field and Magnetic field

Question

A long solenoid having 200 turns per cm carries a current of 1.5 amp. At the can be of it is placed a coil of 100 turns of cross-sectional area 3.14 x 10^-4 m² having its axis parallel to the field produced by the solenoid. When the direction of current in the solenoid is reversed within 0.05 sec, the induced e.m.f. in the coil is

Easy

A

0 48 V
B

0.048 V
C

0.0048 V
D

48 V

Solution

$B = μ_{0} ni \begin{array}{l} = (4 π \times 10^{- 7}) (200 \times 10^{- 2}) \times 1 \cdot 5 \\ = 3 \cdot 8 \times 10^{- 2} W / m^{2} \end{array}$
Magnetic flux through each turn of the coil
$\begin{array}{l} Φ = BA \\ = (3 \cdot 8 \times 10^{- 2}) (3 \cdot 14 \times 10^{- 4} \\ = 1 \cdot 2 \times 10^{- 5} weber \end{array}$
When the current in the solenoid is reversed, the change in magnetic flux
$= 2 \times (1 \cdot 2 \times 10^{- 5}) = 2 \cdot 4 \times 10^{- 5} weber$
Induced e.m.f.
$= N \frac{dΦ}{dt} = 100 \times \frac{2 \cdot 4 \times 10^{- 5}}{0 \cdot 05} = 0 \cdot 048 V$

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