The long, thin wire shown in the figure ( Figure 1 ) is in a region of constant

Question

The long, thin wire shown in the figure (Figure 1) is in a region of constant magnetic field B? . The wire carries a current of 6.2A  and is oriented at an angle of 7.5? to the direction of the magnetic field.

a) If the magnetic force exerted on this wire per meter is 3.5

The long, thin wire shown in the figure (Figure 1) is in a region of constant magnetic field B?. The wire carries a current of 6.2A and is oriented at an angle of 7.5? to the direction of the magnetic field. If the magnetic force exerted on this wire per meter is 3.5times10?2N, what is the magnitude of the magnetic field? At what angle will the force exerted on the wire per meter be equal to 1.4times10?2N?

Explanation / Answer

(a)The magnetic force on the wire per meter is

F = i * B * l * sin? --------(1)

or (F/l) = i * B * sin?

Here,(F/l) = 3.5 * 10-2N,i = 6.2 A and ? =7.5o

or B = [(F/l)/(i * sin?)]

or B = [(3.5 * 10-2)/(6.2 *sin(7.5o))]

or B = [(3.5 * 10-2)/(6.2 * 0.1305)]

or B = 4.326 * 10-2T = 43.26 * 10-3T =43.26 mT

(b)From equation (1),we get

sin? = (F/i * B * l)

or ? = sin-1(F/i * B * l)

Here,(F/l) = 1.4 * 10-2N,i = 6.2 A and B = 43.26mT

or ? = sin-1(1.4 * 10-2/6.2 * 43.26 *10-3)

or ? = sin-1(0.052197) = 2.992o

Therefore,the angle at which the force exerted on the wire permeter be equal to 1.4 * 10-2N is ? =2.992o.

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