s magnctic field passes through a tabletop with a magnitude of So T& directed at

Question

s magnctic field passes through a tabletop with a magnitude of So T& directed at an ange e relative to the upward normal of the tabletop. If the tabletop is 1.5m by 25 magnetic flux through it? (a)07.924-10-5Wh (b) 4.85×10-5 wb (c) 1.699-104 wb (d) 845-W"Wh (e) 7924-W" wb magnitude of 17. Each second, 2.5x101" electrons in a straight wire. Determine the magnitude of the magnetic field at a radius of 7.5 em from the wire. (b) 5.33-10-5T (c) 262×10-1T (d) 4.14-10"t (e) 1.07x lor-- (a) 1.07x10. T 8. The figure shows a uniform, B 1.5 mT magnetic field that is normal to the lane of a conducting, 10 turn circular loop with a resistance of, R 5.32, & a dius of r 2.4 em. The magnetic field is directed out of the paper as shown. ote: The area of the non-circular portion of the wire is considered negligible mpared to that of the circular loop. What is the magnitude of the average induced in the loop if the magnitude ofthe magnetic field is quadrupled in 3 36.19 v (b) 8.64 V (c) 2.036 v (d) 27.1 v (e) 0.0271 v

Explanation / Answer

Ans 16.

Magnitude of magnetic flux is given by formula: Flux = BAcos(theta)

B = magnetic field

A = Area of surface

and theta is angle between the normal to the surface and a magnetic field vector

here

Flux = (50e-6)(1.5*2.5)(cos(65))= 7.924e-5

(a) is correct answer

Ans 17

Current = charge flowing per second

here Q = 2.5e19*1.6e-19 C = 4C

Thus current I = Q/t = 4 A

Magnetic field due to long wire at a given distance d = (u/4pi)(2*I/d) = (10e-7)((2*4)/0.075) = 1.07e-5 T

option (e) is correct

Ans 18

Magnetic field B = 1.5e-3 T

number of turns N = 10

radius of coil R = 0.024 m

EMF induced = N*(dB/dt)*(Area of coil) = 10*(4.5e-3/3e-6)*(3.14*0.024*0.024) = 27.1 V ---- option d

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