Lecture 11 with notes

# Lecture 11 with notes - Magnetic force on a current...

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B F Consider a wire of length which carries a current as shown in the figure. A uniform magnetic field is present in the vicinity of the wire. Experimentaly Li B Magnetic force on a current carrying wire. it was found that a force is exerted by on the wire, and that is perpendicular to the wire. The magnetic force on the wire is the vector sum of all the magnetic forces exerted by on the B B F BF B electrons that constitute . The total charge that flows through the wire in time is given by: Here is the drift velocity of the electrons in the wire. The magnetic force sin9 d d Bd iq t L q it i v v F qv B 0 d d L i v B iLB v B F iLB (28 11)

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An electron travels due north through a vacuum in a region of uniform magnetic Feld B that is also directed due north. The electron will: 1 2 3 4 5 0% 0% 0% 0% 0% 1. be unaffected by the field 2. speed up 3. slow down 4. follow a right- handed helical path 5. follow a left-handed helical path
. dL B dF i If we assume the more general case for which the magnetic field froms and angle with the wire the magnetic force equation can be writ B Magnetic force on a straight wire in a uniform magnetic field. ten in vector form as: Here is a vector whose magnitude is equal to the wire length and has a direction that coincides with that of the current. The magnetic force magnitude B B F iL B L L F sin In this case we divide the wire into elements of length which can be considered as straight. The magnetic f iLB dL Magnetic force on a wire of arbitrary shape placed in a non -uniform magnetic field. orce on each element is: The net magnetic force on the wire is given by the integral: B B dF idL B F i dL B = B F iL B B dF idL B = B F i dL B (28 12)

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C C Top view Side view Consider the rectangular loop in fig.a with sides of lengths and which carries ˆ a current . The loop is placed in a magnetic field so that the normal to the lo ab in Magnetic torque on a current loop 13 24 op forms an angle with . The magnitude of the magnetic force on sides 1 and 3 is: sin90 . The magnetic force on sides 2 and 4 is: sin(90 ) cos . These forces cancel i B F F iaB iaB F F ibB ibB n pairs and thus 0 The torque about the loop center C of and is zero because both forces pass through point C. The moment arm for and is equal to ( / 2)sin . The two torques tend to ro net F FF F F b tare the loop in the same (clockwise) direction and thus add up. The net torque + =( / 2)sin ( sin sin iabB iabB iabB iAB sin net iAB 0 net F (28 13)
A square loop of wire lies in the plane of the page and carries a current I as shown. There is a uniform magnetic feld B parallel to the side MK as indicated. The loop will tend to rotate: 1 2 3 4 0% 0% 0% 0% 1. About PQ with KL coming out of the page 2. About PQ with MN coming our of the page 3. About SR with LN out of the page 4. About SR with MN out of the page

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UB The torque of a coil that has loops exerted by a uniform magnetic field and carrries a current is given by the equation: We define a new vector associated wi N B i NiAB Magnetic dipole moment : th the coil
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Lecture 11 with notes - Magnetic force on a current...

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