The calculated mutual inductance is 8.544 x 10⁻⁵ H.
Two coils have a mutual inductance of 1 henry when emf of 1 volt is induced in coil 1 and when the current flowing through coil 2 is changing at the rate of one ampere per second.
Length of the solenoid= 5.0 cm
Area of cross-section=1.0 cm²
no of spaced turns=300 turns
turns of insulated wire=180 turns
Mutual inductance (M) = μ₀μr N1N2 A/ L
=(4xπx 10⁻⁷) x (6.3 x 10⁻³) x 300 x 180 x 1/ 5
=79.12 x 10⁻¹⁰ x 54000 / 5
=8.544 x 10⁻⁵ H
hence, the mutual inductance is 8.544 x 10⁻⁵ H.
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Answer:200J
Explanation:E=MGH
= 2*10(assuming you round earth’s gravity to 10)*10
=200
Answer:
94.67 N
Explanation:
Consider a free body diagram with force, F of 41 N applied at an angle of 37 degrees while the weight acts downwards. Resolving the force into vertical and horizontal components, we obtain a free body diagram attached.
At equilibrium, normal reaction is equal to the sum of the weight and the vertical component of the force applied. Applying the condition of equilibrium along the vertical direction.

Substituting 70 N for W, 41 N for F and
for 37 degrees
N=70+41sin37=94.67441595 N and rounding off to 2 decimal places
N=94.67 N
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