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Chapter 28: Q40P (page 831)

A wire $1.80$ m long carries a current of $13.0$ A and makes an angle of $35.0 掳$ with a uniform magnetic field of magnitude $B = 1.50$ T. Calculate the magnetic force on the wire.

Short Answer

Expert verified

The magnetic force on the wire is $20.1 N$ .

Step by step solution

01

Listing the given quantities

Current in the conductor, $i = 13.0 A$
Magnetic field, $B = 1.50 T$
Length of the conductor, $L = 1.80 m$
The angle between current and field, $胃 = 35 掳$

02

Understanding the concept of magnetic force

By using the formulas of magnetic force in terms of current and magnetic field and substituting the given quantities, we can find the magnitude of the magnetic force.

Formula:

Magnetic force, $F_{B} = i L B s i n 胃$

03

Calculation of magnetic force on the wire

The magnetic force is given by:

$F_{B} = i L B s i n 胃$

Where

$i = C u r r e n t, L = L e n g t h o f t h e c o n d u c t o r, B = M a g n e t i c f i e l d, 胃 = A n g l e b e t w e e n c u r r e n t a n d f i e l d$

Therefore,

role="math" localid="1662458941501" $F_{B} = (13.0) (1.80) (1.50) (\sin 35 掳) = 20.1 N$

The magnetic force on the wire is $20.1 N$ .

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Most popular questions from this chapter

An electron that has velocity $\overset{鈫�}{谓}$ =( $2.0 脳 10^{6}$ m/s) $\hat{i}$ + (3.0 $脳 10^{6}$ m/s)moves through the uniform magnetic field $\overset{鈫�}{B}$ = $(0.030 T) \hat{i} - (0.15 T) \hat{j}$ .(a)Find the force on the electron due to the magnetic field. (b)Repeat your calculation for a proton having the same velocity.

Bainbridge鈥檚 mass spectrometer, shown in Fig. 28-54, separates ions having the same velocity. The ions, after entering through slits, S₁and S₂, pass through a velocity selector composed of an electric field produced by the charged plates Pand P', and a magnetic field $\overset{鈫�}{B}$ perpendicular to the electric field and the ion path. The ions that then pass undeviated through the crossed $\overset{鈫�}{E}$ and $\overset{鈫�}{B}$ fields enter into a region where a second magnetic field $\overset{鈫�}{B}$ exists, where they are made to follow circular paths. A photographic plate (or a modern detector) registers their arrival. Show that, for the ions, q/m=E/rBB,where ris the radius of the circular orbit.

An alpha particle travels at a velocity $\overset{鈫�}{v}$ of magnitude $550$ m/s through a uniform magnetic field of magnitude $\overset{鈫�}{B}$ =0.045T. (An alpha particle has a charge of ${+3.2脳10}^{-19}$ C and a mass ofkg.) The angle between $\overset{鈫�}{v}$ and $\overset{鈫�}{B}$ is 52掳. (a)What is the magnitude of the force $\overset{鈫�}{F_{B}}$ acting on the particle due to the field? (b)What is the acceleration of the particle due to $\overset{鈫�}{F_{B}}$ ?

(c)Does the speed of the particle increase, decrease, or remain the same?

Question: A proton moves through a uniform magnetic field given by $\overset{鈫�}{B} = (10 \hat{i} + 20 \hat{j} + 30 \hat{k}) mT$ . At time $t_{1}$ , the proton has a velocity given $\overset{鈫�}{v} = v_{x} \hat{i} + v_{y} \hat{j} + (2.00 km / s) \hat{k}$ and the magnetic force on the proton is $\overset{鈫�}{F_{B}} = (4.0 脳 10^{- 17} N) \hat{i} + (2.0 脳 10^{- 17} N) \hat{j}$ .At that instant, (a) What is the value of $v_{x}$ ? (b)What is the value of $v_{y}$ ?

An electron is moving at $7.20 脳 10^{6} m / s$ in a magnetic field of strength 83.0mT. What is the (a) maximum and (b) minimum magnitude of the force acting on the electron due to the field? (c) At one point the electron has an acceleration ofmagnitude role="math" localid="1662987933736" $4.90 脳 10^{14} m / s^{2}$ .What is the angle between the electron鈥檚 velocity and the magnetic field?

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