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ladessa [460]
4 years ago
14

1. A uniform magnetic field is directed vertically upwards. In which direction in this field should an alpha particle be project

ed so that it is deflected south ward? Name and state the rule you have used to find the direction in this case.
Physics
1 answer:
max2010maxim [7]4 years ago
3 0

Answer:

1. Fleming's left hand rule

2. It must be projected towards the east

Explanation:

Fleming's left-hand rule states that; When a current-carrying conductor is placed in an external magnetic field, the conductor experiences a force perpendicular to both the field and to the direction of the current flow. This rule was first put forward by John Ambrose Fleming in the later part of the nineteenth century.

Hence if the thumb, fore finger and middle finger of the lefthand are held mutually at right angles to each other; the thumb shows the direction of motion, the fore finger shows the direction of the field while the middle finger shows the direction of the current.

Hence, if the alpha particle is projected eastwards(at right angles) to the uniform magnetic field, it will be deflected southwards in the magnetic field.

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3.2 m

Explanation:

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1 year ago
Determine a formula for the magnitude of the force F exerted on the large block (Mc) so that the mass Ma does not move relative
SVEN [57.7K]

Answer:

The magnitude of the force F is given by

F =  (M_{a} + M_{b} + M_{c} ) *(M_{b}*g/(\sqrt{M_{a} ^{2}-M_{b} ^{2}}))

Explanation:

Given there are three blocks of masses M_{a}, M_{b} and M_{c} (ref image in attachment)

When all three masses move together at an acceleration a, the force F is given by

F =  (M_{a} + M_{b} + M_{c} ) *a    ................(equation 1)

Also it is given that M_{a} does not move with respect to M_{c}, which gives tension T  is exerted on pulley  by M_{a} only, Hence tension T is

T = M_{a} *a    ..........(equation 2)

There is also also tension exerted by M_{b}. There are two components here: horizontal due to acceleration a and vertical component due to gravity g. Thus tension is given by

T = M_{b} \sqrt{a^{2} +g^{2} }   ................(equation 3)

From equation 2 and 3, we get

M_{a} *a  = M_{b} \sqrt{a^{2} +g^{2} }  

Squaring both sides we get

M_{a} ^{2} *a^{2} = M_{b} ^{2} * (a^{2}+g^{2})

M_{a} ^{2} *a^{2} = (M_{b} ^{2} * a^{2})+ (M_{b} ^{2} *g^{2})

(M_{a} ^{2}  -  M_{b} ^{2}) * a^{2} = M_{b} ^{2} *g^{2}

a^{2} = M_{b} ^{2} *g^{2}/(M_{a} ^{2}  -  M_{b} ^{2})

Taking square root on both sides, we get acceleration a

a = M_{b}*g/(\sqrt{M_{a} ^{2}-M_{b} ^{2}})

Hence substituting the value of a in equation 1, we get

F =  (M_{a} + M_{b} + M_{c} ) *(M_{b}*g/(\sqrt{M_{a} ^{2}-M_{b} ^{2}}))

3 0
3 years ago
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Sholpan [36]

Answer:

The acceleration of the body, a = 2193 m/s²

Explanation:

Given,

The mass of the body, m = 0.3 kg

The force acting on the body, F = 657.9 N

The force acting on an object is proportional to the product of mass and acceleration of the body.

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Therefore, the acceleration of the body is

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                              = 657.9 N / 0.3 kg

                              = 2193 m/s²

Hence, the acceleration of the body, a = 2193 m/s²

4 0
4 years ago
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