A coil of wire with 50 turns and a cross-sectional area of 0.02 m^2 lies with it

Question

A coil of wire with 50 turns and a cross-sectional area of 0.02 m^2 lies with its plane perpendicular to a magnetic field of 2 T. The coil is rapidly removed from the field in a time of 0.5 s. What is the initial magnetic flux through the coil? Show calculation: __________ Tm^2 What is the induced voltage in the coil? __________ ___ A particle with a charge of 0.06 C and velocity 1000 m/s is moving perpendicular to a uniform with a strength of 0.5 T. What is the magnitude of the force exerted on the particle?

Explanation / Answer

Question 17) Magnetic force is given by formula

Fb = q(V x B) = qVBsin(©)

Here © is the angle , angle(©) = 90° since the particle is moving perpendicular to magnetic field

Given , B = 0.5 T, V = 1000m/s , q = 0.06 C

So, Fb = 0.06*1000*0.5*sin(90°)

Fb = 30 N

So, magnetic force = 30 N

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