Question

(a) An electron is accelerated through an electron gun to a speed of 2 × 10 ? m/s. The electron is shot perpendicularly into a region of magnetic field of B = 2 × 10?? T. It was observed that the electron is traveling in a circular path within the region. Calculate the radius of the path of the electron. (b) An electron travels at a velocity of v? = 2.5 × 10? i? m/s through a uniform magnetic field of B? = 0.05 i? T. (i) Calculate the magnitude of the magnetic force acting on the electron due to the field. (ii) Calculate the acceleration of the electron due to the magnetic force. (c) Two long straight wires are parallel and 8 cm apart. The magnitude of the current in the wires are 0.5 A and 0.35 A respectively. Given that the current are in the same direction and magnetic force acting on each other. (i) Calculate the magnitude of the force per unit length acting on each wire due to the other wire. (ii) Is the force attractive or repulsive?

          (a) An electron is accelerated through an electron gun to a speed of 2 × 10 ? m/s. The electron is shot perpendicularly into a region of magnetic field of B = 2 × 10?? T. It was observed that the electron is traveling in a circular path within the region. Calculate the radius of the path of the electron.
(b) An electron travels at a velocity of v? = 2.5 × 10? i? m/s through a uniform magnetic field of B? = 0.05 i? T.
(i) Calculate the magnitude of the magnetic force acting on the electron due to the field.
(ii) Calculate the acceleration of the electron due to the magnetic force.
(c) Two long straight wires are parallel and 8 cm apart. The magnitude of the current in the wires are 0.5 A and 0.35 A respectively. Given that the current are in the same direction and magnetic force acting on each other.
(i) Calculate the magnitude of the force per unit length acting on each wire due to the other wire.
(ii) Is the force attractive or repulsive?

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(a) An electron is accelerated through an electron gun to a speed of 2 × 10 ? m/s. The electron is shot perpendicularly into a region of magnetic field of B = 2 × 10?? T. It was observed that the electron is traveling in a circular path within the region. Calculate the radius of the path of the electron.
(b) An electron travels at a velocity of v? = 2.5 × 10? i? m/s through a uniform magnetic field of B? = 0.05 i? T.
(i) Calculate the magnitude of the magnetic force acting on the electron due to the field.
(ii) Calculate the acceleration of the electron due to the magnetic force.
(c) Two long straight wires are parallel and 8 cm apart. The magnitude of the current in the wires are 0.5 A and 0.35 A respectively. Given that the current are in the same direction and magnetic force acting on each other.
(i) Calculate the magnitude of the force per unit length acting on each wire due to the other wire.
(ii) Is the force attractive or repulsive?

Added by Audrey R.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Madhur L

06/02/2022

Step 1

Given: - Mass of electron, \( m = 9.1 \times 10^{-31} \) kg - Speed of electron, \( v = 2 \times 10^{6} \) m/s - Charge of electron, \( Q = 1.6 \times 10^{-19} \) C - Magnetic field, \( B = 2 \times 10^{-4} \) T Plugging in the values: \[ R = \frac{(9.1 \times Show more…

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(a) An electron is accelerated through an electron gun to a speed of 2 x 10^6 m/s. The electron is shot perpendicularly into a region of magnetic field of B = 2 x 10^-4 T. It was observed that the electron is traveling in a circular path within the region. Calculate the radius of the path of the electron. (b) An electron travels at a velocity of v = 2.5 x 10^5 i m/s through a uniform magnetic field of B = 0.05 i T. (i) Calculate the magnitude of the magnetic force acting on the electron due to the field. (ii) Calculate the acceleration of the electron due to the magnetic force. (c) Two long straight wires are parallel and 8 cm apart. The magnitude of the current in the wires are 0.5 A and 0.35 A respectively. Given that the current are in the same direction and magnetic force acting on each other. (i) Calculate the magnitude of the force per unit length acting on each wire due to the other wire. (ii) Is the force attractive or repulsive?

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Transcript

00:01 Hello student, to solve this question, let us write the formula for magnetic force fb, which is equal to qv vector cross b vector.

00:09 It can be also written as qvb when there is theta equals 90 degree.

00:15 Also, the centripetal force fc can be written as mv squared divided by r.

00:21 Now from here, firstly, let us calculate part a, which is that is radius of path of the electron.

00:28 So using the centripetal force, here, here, here which will be equal to the magnetic force.

00:33 So the relation will become mv squared divided by r equals qu vb.

00:40 So from here we can find the value of r which will be equal to mv divided by qb.

00:48 So putting the values given in the statement we get mass of an electron is 9 .1 into 10 to power minus 31.

00:56 Multiply by speed is given 2 into 10 to power 6.

01:00 Meter per second divided by charge on electron is 1 .6 into 10 to power minus 19 multiplied by magnetic field is given 2 into 10 to power minus 4 tesla.

01:13 So from here we can find the answer for radius which will be 5 .69 multiplied by 10 to power minus 2.

01:22 So the final answer for part a will be radius which is equal to 5 .69 multiplied by 10 to power minus 2 meters.

01:32 Now for part b we need to firstly calculate the magnetic force for part 1 which can be calculated using the same results so putting here fb equals q is given in that is 1 .6 into 10 as 10 minus 19 multiplied by v vector is given as 2 .5 multiply by 10 as to power 5 j then cross product with the b vector which is given as 0 .05i.

02:04 Now after solving this relation we get the magnetic force vector that is fb vector will be equal to minus 0 .2k unit vector newton.

02:18 So the final answer that is the magnitude of fb from here will be equal to fb is 0 .2 newton.

02:27 Now we need to calculate the acceleration for part 2, which will be given as acceleration a will be equal to force fb divided by mass m...

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