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PhysicsEMI/ACMiscellaneousEasy2 minPYQ_2020

PhysicsEasynumerical

In a magnetic field of $0.05 T$ , area of a coil changes from $101 {cm}^{2}$ to $100 {cm}^{2}$ without changing the resistance which is $2 Ω$ . The amount of charge (in $μ C$ ) that flows during this period is

Answer:

2.50

Solution:

$B = 0.05 T$

$A_{1} = 101 {cm}^{2} = 101 \times 10^{- 4} m^{2}$

$A_{2} = 100 {cm}^{2} = 100 \times 10^{- 4} m^{2}$

$R = 2 Ω$

Amount of charge flown is given by,

$q = \frac{B (Δ A)}{R}$

$= \frac{0.05 \times (101 \times 1 0^{- 4} - 100 \times 1 0^{- 4})}{2}$

$= \frac{0.05 \times 1 \times 1 0^{- 4}}{2}$

$= 2.5 \times 1 0^{- 6} C$

Stream:JEESubject:PhysicsTopic:EMI/ACSubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2020

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