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Question

Physics Question on Inductance

Two coils are at fixed locations. When coil 1 has no current and the current in coil 2 increases at the rate of 15.0 A/s, the emf in coil 1 is 25.0 mV, when coil 2 has no current and coll 1 has a current of 3.6 A, the flux linkage in coil 2 is

A

16 mWb

B

10 mWb

C

4.00 mWb

D

6.00 mWb

Answer

6.00 mWb

Explanation

Solution

Coefficient of mutual inductance M is given by ∣M∣=e1(di2/dt)=ϕ2i1|M| = \frac{e_1}{(di_2 /dt)} = \frac{\phi_2}{i_1} ∴ ϕ2=e1i1(di2/dt)=(25.0×10−3)(3.6)(15)\therefore \, \phi_2 = \frac{e_1 i_1}{(di_2 /dt)} = \frac{(25.0 \times 10^{-3})(3.6)}{(15)}