For each question, circle the letter corresponding to the correct answer. A mtal

Question

For each question, circle the letter corresponding to the correct answer. A mtallic rod moves at constant speed in the positive x direction inside a uniform magnetic fiels as shown in the figure below. Positive and negative charges build up on the rod as indicated. What is the direction of the magnetic field? positive y positive z negative x negative y negative z two charges, each having a mangitude of 7.0 nC with one positive and negative, are located on the x axis at x = plusminus 1.2m. What is the magnitude of the total electric field on the y axis at y = 0.50 m? 29 N/C 75 N/C 69 N/C 34 N/C 0 The electric potential of a charge distribution is given by the equation V(x) = 3x2y2 + yz3 - 2z3x where x,y,z are measured in meters and V is measured in volts. Calculated the magnitude of the electric field vector at the position(x,y,z) = (1,1,1) 74V/m 2.0 v/m 4.3 V/m -8.1V/m 8.6 V/m

Explanation / Answer

1). answer is E

since F=qVXB

so on proton force is in +y direction so by right hand rule magnetic field must be in -ve z direction

2). we have

tan theta= 0.5/1.2

so theta=22.6 degree

now

r=sqrt(0.5^2+1.2^2)

=1.3 m

so

EF= 2Esin theta

=2sin22.6* KQ/r^2

=29N/c

ans( A)

3). Ex= -dv/dx= -6xy^2 +2z^3 = -4i

Ey= -dv/dy = -6x^2y -z^3 = -7k^

Ez=-dv/dz= -3yz^2 +6z^2x= 3k^

so magnitude is

sqrt(3^2+7^2+4^2)

8.6 V/m

answer

E.

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