A current-carrying wire is bent into a circular loop of radius R and lies in an xy plane. A uniform external magnetic field B in the +z direction exists throughout the plane of the loop. The current has the magnitude of I and it is deirected counterclockwise when observing from positive z axis.What is the magnetic force exerted by the external field on the loop?Express your answer in terms of some or all of the variables I, R, and B

Answers

Answer 1

Answer:

Final answer:

The net magnetic force exerted by the external magnetic field on a current-carrying wire formed into a loop in a uniform magnetic field is absolutely zero since the individual forces on each section of the loop cancel each other out.

Explanation:

The force exerted by a magnetic field on a current carrying wire is given by Lorentz force law, which says that the force is equal to the cross product of the current and the magnetic field. However, in this case, where the wire is formed into a loop with current flowing in a counter-clockwise direction in presence of an external magnetic field, the individual forces on each infinitesimal section of the loop cancel each other out. Therefore, the net magnetic force exerted by the external field on the entire loop is zero.

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Answer 2

Answer:

Final answer:

The magnetic force exerted on a current-carrying wire loop by an external magnetic field can be calculated using the equation F = I * R * B.

Explanation:

The magnetic force exerted by the external field on the current-carrying wire loop can be determined using the equation F = I * R * B. The magnetic force is equal to the product of the current, radius, and magnetic field strength. The direction of the magnetic force can be determined using the right-hand rule, where the thumb represents the direction of the current, the fingers represent the magnetic field, and the palm represents the direction of the force.

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A popular physics lab involves a hand generator and an assortment of wires with different values of resistance. In the lab, the leads of the generator are connected across each wire in turn. For each wire, students attempt to turn the generator handle at the same constant rate. Students must push harder on the handle when the leads of the generator are connected__________. This is because turning the handle at a given constant rate produces__________ , regardless of what is connected to the leads. So, when turning the handle at a constant rate, lab students must push harder in cases where there is________

Answers

Answer:

Explanation:

Students must push harder on the handle when the leads of the generator are connected across the wire with the lowest resistance.

This is because turning the handle at a given constant rate produces a constant voltage across the leads, regardless of what is connected to the leads.

So, when turning the handle at a constant rate, lab students must push harder in case where there is a greater current through the connected wire.

A 79-turn, 16.035-cm-diameter coil is at rest in a horizontal plane. A uniform magnetic field 43 degrees away from vertical increases from 0.997 T to 6.683 T in 56.691 s. Determine the emf induced in the coil.

Answers

Answer:

The induced emf is 0.0888 V.

Explanation:

Given that,

Number of turns = 79

Diameter = 16.035 cm

Angle = 43

Change in magnetic field $\Delta B=(6.683-0.997)= 5.686\ T$

Time = 56.691 s

We need to calculate the induced emf

Using formula of induced emf

$\epsilon=(NA\Delta B\cos\theta)/(\Delta T)$

Where, N = number of turns

A = area

B = magnetic field

Put the value into the formula

$\epsilon=(79*\pi*(8.0175*10^(-2))^2*5.686*\cos43)/(56.691)$

$\epsilon =0.0888\ V$

Hence, The induced emf is 0.0888 V.

The uniform crate has a mass of 50 kg and rests on the cart having an inclined surface. Determine the smallest acceleration that will cause the crate either to tip or slip relative to the cart. What is the magnitude of this acceleration

Answers

Answer:

The answer is below

Explanation:

Let g = acceleration due to gravity = 9.81 m/s², x = half of the width of the crate, half of the height of the crate = 0.5 m, a = acceleration of crate, N = force raising the crate

The sum of moment is given as:

$50asin(15)x+50acos(15)0.5=-50(9.81)sin(15)0.5+50(9.81)cos(15)x\ \ \ (1)$

Sum of vertical forces is zero, hence:

$N-50(9.81)cos(15)+50acos(15)=0\ \ \ (2)$

Sum of horizontal force is zero, hence:

$50(9.81)sin(15)-\mu N+50acos(15)=0\n\n50(9.81)sin(15)-0.5 N+50acos(15)=0\ \ \ (3)$

Solving equation 1, 2 and 3 simultaneously gives :

N = 447.8 N, a = 2.01 m/s², x = 0.25 m

x is supposed to be 0.3 m (0.6/2)

The crate would slip because x <0.3 m

A laser (electromagnetic wave) has the maximum electric field strength of 1.0x1011 V/m. What is the force the laser applies on a mirror (totally reflective) of 5.0 mm2 area? A. 2.76 x105N B. 1.21 x106N C. 1.94 x106N D.4.43 x105 N E. 7.82 x104N

Answers

Answer:

The correct option is D

Explanation:

From the question we are told that

The maximum electric field strength is $E = 1.0 *10^(11) \ V/m$

The area is $A = 5.0 \ mm^2 = 5.0 *10^(-6) \ m^2$

Generally the force the laser applies is mathematically represented as

$F = \epsilon_o * E ^2 * A$

Here $\epsilon_o = 8.85*10^(-12) C/(V \cdot m)$

$F = 8.85*10^(-12) * (1.0 *10^(11)) ^2 * 5.00*10^(-6 )$

=> $F = 4.43 *10^(5) \ N$

A 8.00-μF capacitor that is initially uncharged is connected in series with a 3.00-Ω resistor and an emf source with E = 70.0 V and negligible internal resistance. At the instant when the resistor is dissipating electrical energy at a rate of 300 W, how much energy has been stored in the capacitor?

Answers

Answer:

The energy stored is $E_s = 0.0064 \ J$

Explanation:

From the question we are told that

The capacitance is $C = 8 \ \mu F = 8*10^(-6) \ F$

The resistance is R = 3.00-Ω

The emf is $E_t = 70.0 V$

The power is P = 300 W

Generally the total emf is mathematically represented as

$E_t = E_c + E_r$

Here $E_c$ is the emf across that capacitor which is mathematically represented as

$E_c = (q)/(C)$

and $E_r$ is the emf across the resistor which is mathematically represented as

$E_r = √(P R)$

$E_t = √(PR) + (q)/(C)$

=> $q = C[E_t - √(PR) ]$

Generally the energy stored in a capacitor is mathematically represented as

$E_s = (q^2)/(2C)$

=> $E_s = ([C [ E_t - √(PR) ]]^2)/(2C)$

=> $E_s = ([8.0*10^(-6) [ 70 - √(300 * 3))/(2 *(8.0*10^(-6)))$

=> $E_s = 0.0064 \ J$

Final answer:

The energy stored in the capacitor is 0 J.

Explanation:

When a 8.00-μF capacitor that is initially uncharged is connected in series with a 3.00-Ω resistor and an emf source with E = 70.0 V

At the instant when the resistor is dissipating electrical energy at a rate of 300 W, we can calculate the current flowing through the circuit using Ohm's law: I = V/R = 70.0 V / 3.00 Ω = 23.33 A.

The energy stored in a capacitor can be calculated using the formula: E = 1/2 * C * V^2, where C is the capacitance and V is the voltage across the capacitor.

Since the capacitor is initially uncharged, the voltage across it is also zero. So the energy stored in the capacitor is 0.5 * 8.00 x 10^-6 F * (0 V)^2 = 0 J.

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Patricia places 1500g of fruit on a scale compressing it by 0.02m. What is the force constant of the spring on the scale. Hint: you need 2 equations to solve this.