# B suppose now that the bar is held fi xed in the

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b. Suppose now that the bar is held fi xed in the position shown and that the magnetic fi eld now increases with time as B = 0 . 8 t 2 Tesla. What is the magnitude and direction of the induced current in the closed loop at t = 4 sec? Now the area of the loop is constant but the magnetic fi eld is changing. The induced emf is now given by ε = d Φ B dt = A dB dt = ( l )( x ) d dt (0 . 8 t 2 ) = ( l )( x )(1 . 6) t = (2 . 5)(5)(1 . 6)(4) = 80 V olts The current is then given by I = ε R = 80 12 = 6 . 67 Amps Since the fl ux through the loop is increasing into the paper due to the increasing magnetic fi eld, the fl ux that will be generated by the current in the loop needs to be out of the paper. To get this fl ux out of the paper, the induced current will need to be counterclockwise . c. Suppose now that the fi eld is increasing with time as B = 0 . 8 t 2 Tesla and also that the bar is moving to the left with a velcoity of 1 4 m/sec. What is the magnitude and direction of the current in the loop at t = 0 . 24 sec? Now both the area and the magnetic fi eld are changing. The induced emf is now given by ε = d Φ B dt = ( B dA dt + A dB dt ) The area that is used for the second term is given by A = l ( x 0 + vt ) 2

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We then have for the emf ε = d Φ B dt = ( B dA dt + A dB dt ) = (0 . 8 t 2 l v + [ l ( x 0 + vt )](1 . 6) t ) = ((0 . 8)(0 . 24) 2 (2 . 5)( 0 . 25) + [2 . 5(5 (0 . 25)(0 . 24))](1 . 6)(0 . 24)) = (( 0 . 0288) + [12 . 35](0 . 384)) = 4 . 7136 V olts The current is given by I = ε R = 4 . 7136 12 = 0 . 3928 Amps Since the fl ux through the loop is increasing into the paper due to the net e ff ect of increasing magnetic fi eld and decreasing area, the fl ux that will be generated by the current in the loop needs to be out of the paper. To get this fl ux out of the paper, the induced current will need to be counterclockwise . 3
Soc. Sec # Name 3. A particle carries a charge of 2 coulombs. When it moves with a velocity of 3 m/s at an angle above the x-axis on the xy plane, a uniform magnetic fi eld exerts a force on the particle in the negative y direction. When the same particle moves with a velocity of 2 m/s along the positive z-axis, there is a force of 4 Newtons exerted on the particle in the positive x-direction. What are the magnitude and direction of the magnetic fi eld? We remember that −→ F = q −→ v × −→ B We can write this as −→ F = q ¯ ¯ ¯ ¯ ¯ ¯ b i b j b k v x v y v z B x B y B z ¯ ¯ ¯ ¯ ¯ ¯ −→ F = q h b i ( v y B z v z B y ) + b j ( v z B x v x B z ) + b k ( v x B y v y B x ) i (1) We start with the fi rst situation: Since the force is only in the negative y-direction only the second term of equation 1 comes into play.

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• Spring '08
• Bingham
• Physics, Magnetic Field, ε

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